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1. Shard Calculation

Query:

Response:

Calculated Shard: 185 (from laksa154)

2. Crawled Status Check

Query:

curl -X POST \
  'http://laksa185.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://byjus.com/physics/relation-between-electric-field-and-electric-potential/\')), getAhrefsUnparsedNoserviceFromURL(\'https://byjus.com/physics/relation-between-electric-field-and-electric-potential/\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/byjus.com\/physics\/relation-between-electric-field-and-electric-potential\/","crawl_time":1773998448,"first_indexed_time":1553056775,"http_code":200,"src_unparsed":"com,byjus!\/physics\/relation-between-electric-field-and-electric-potential\/ s443","src_root_hash":"6322170921519322385","history_drop_reason":null,"meta_title":"","meta_descriptions":[],"attrs_boilerpipe_text":"The electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as –\n“Electric field is the negative space derivative of electric potential.”\nElectric Field and Electric Potential\nThe relation between the electric field and\nelectric potential\nis mathematically given by\nE\n=\n−\nd\nV\nd\nx\nWhere,\nE is the Electric field.\nV is the electric potential.\ndx is the path length.\n– sign indicates that the electric field is directed from higher potential to lower potential\nVideo Explanation:\n1,857\nDirection of Electric Field\nIf the field is directed from lower potential to higher then the direction is taken to be positive.\nIf the field is directed from higher potential to lower potential then the direction is taken as negative.\nElectric Field and Electric Potential Relation\nTest charge\nFormula\nElectric gradient\nPositive\nw\nq\n0\n=\n∫\na\nb\nE\n→\n.\nd\nl\n→\n=\nV\nb\n−\nV\na\nHigher as you go closer towards test charge.\nNegative\nw\nq\n0\n=\n∫\na\nb\nE\n→\n.\nd\nl\n→\n=\nV\na\n−\nV\nb\nHigher as you go move away from test charge.\nEquipotential surface\nw\nq\n0\n=\n∫\na\nb\nE\n→\n.\nd\nl\n→\n=\n0\nElectric potential is perpendicular to Electric field lines.\nElectric Field and Electric Potential Relation Derivation\nLet us study how to find the electric potential of the electric field is given. Work done in moving the test\ncharge\nq\n0\nfrom a to b is given by-\nW\n(\nq\n0\na\n→\nb\n)\n=\n∫\na\nb\nF\n→\n.\nd\nl\n→\n=\nq\n0\n∫\na\nb\nE\n→\n.\nd\nl\n→\nWhere,\nF is the force applied\ndl is the short element of the path while moving it from a to b.\nThe force can be written as charge times electric field\n.\n=\nq\n0\n∫\na\nb\nE\n→\n.\nd\nl\n→\nDividing both sides by test charge q\n0\nw\nq\n0\n=\n∫\na\nb\nE\n→\n.\nd\nl\n→\nWork done by the test charge is the potential V\na\n-V\nb\n∫\na\nb\nE\n→\n.\nd\nl\n→\n=\nV\na\n−\nV\nb\nFor equipotential surface, V\na\n=V\nb\nthus,\n∫\na\nb\nE\n→\n.\nd\nl\n→\n=\n0\nHope you understood the relation and conversion between Electric Field and Electric potential.\nPhysics Related Topics:\nStay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.","attrs_markdown":"![](https:\/\/www.facebook.com\/tr?id=506170199519639&ev=PageView&noscript=1)\n\nThe electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – **“Electric field is the negative space derivative of electric potential.”**\n\n## Electric Field and Electric Potential\nThe relation between the electric field and [electric potential](https:\/\/byjus.com\/jee\/electric-potential-energy\/) is mathematically given by\n\n|   |\n|---|\n| E \\= − d V d x |\n\nWhere,\n\nE is the Electric field.\n\nV is the electric potential.\n\ndx is the path length.\n\n– sign indicates that the electric field is directed from higher potential to lower potential\n\n### Video Explanation:\n![](https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2021\/06\/Relation-between-Electric-Field-and-Potential.jpg)\n\n1,857\n\n### Direction of Electric Field\n- If the field is directed from lower potential to higher then the direction is taken to be positive.\n- If the field is directed from higher potential to lower potential then the direction is taken as negative.\n## Electric Field and Electric Potential Relation\n|   |   |   |\n|---|---|---|\n| **Test charge** | **Formula** | **Electric gradient** |\n| **Positive** | w q 0 \\= ∫ a b E → . d l → \\= V b − V a | Higher as you go closer towards test charge. |\n| **Negative** | w q 0 \\= ∫ a b E → . d l → \\= V a − V b | Higher as you go move away from test charge. |\n| **Equipotential surface** | w q 0 \\= ∫ a b E → . d l → \\= 0 | Electric potential is perpendicular to Electric field lines. |\n## Electric Field and Electric Potential Relation Derivation\nLet us study how to find the electric potential of the electric field is given. Work done in moving the test [charge](https:\/\/byjus.com\/physics\/electric-charge\/) q0 from a to b is given by-\n\nW\n\n(\n\nq\n\n0\n\na\n\n→\n\nb\n\n)\n\n\\=\n\n∫\n\na\n\nb\n\nF\n\n→\n\n.\n\nd\n\nl\n\n→\n\n\\=\n\nq\n\n0\n\n∫\n\na\n\nb\n\nE\n\n→\n\n.\n\nd\n\nl\n\n→\n\nWhere,\n\n- F is the force applied\n- dl is the short element of the path while moving it from a to b.\n\n**The force can be written as charge times electric field**.\n\n\\=\n\nq\n\n0\n\n∫\n\na\n\nb\n\nE\n\n→\n\n.\n\nd\n\nl\n\n→\n\nDividing both sides by test charge q0\n\nw\n\nq\n\n0\n\n\\=\n\n∫\n\na\n\nb\n\nE\n\n→\n\n.\n\nd\n\nl\n\n→\n\nWork done by the test charge is the potential Va\\-Vb\n\n∫\n\na\n\nb\n\nE\n\n→\n\n.\n\nd\n\nl\n\n→\n\n\\=\n\nV\n\na\n\n−\n\nV\n\nb\n\n**For equipotential surface, Va\\=Vb** thus,\n\n∫\n\na\n\nb\n\nE\n\n→\n\n.\n\nd\n\nl\n\n→\n\n\\=\n\n0\n\nHope you understood the relation and conversion between Electric Field and Electric potential.\n\n**Physics Related Topics:**\n\n|   |\n|---|\n| [Electric Field Lines](https:\/\/byjus.com\/physics\/electric-field-lines\/) |\n| [Electric Force](https:\/\/byjus.com\/physics\/electrical-force\/) |\n| [Electrolytic Capacitor](https:\/\/byjus.com\/physics\/electrolytic-capacitor\/) |\n| [Electric Circuit and Electrical Symbols](https:\/\/byjus.com\/physics\/electric-circuit-electrical-symbols\/) |\n\n*Stay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.*\n\nTest Your Knowledge On Relation Between Electric Field And Electric Potential\\!\n\nQ5\n\n![Quiz Image](https:\/\/cdn1.byjus.com\/byjusweb\/img\/interactive-quiz\/Quiz_cartoon.png)\n\nPut your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin\\!\n\nSelect the correct answer and click on the “Finish” button  \nCheck your score and answers at the end of the quiz\n\nStart Quiz\n\nCongrats\\!\n\nVisit BYJU’S for all Physics related queries and study materials\n\nYour result is as below\n\n0 out of 0 arewrong\n\n0 out of 0 are correct\n\n0 out of 0 are Unattempted\n\nView Quiz Answers and Analysis\n\nX\n\nLogin To View Results\n\nMobile Number\\*\n\nSend OTP\n\nDid not receive OTP?\n\nRequest OTP on Voice Call\n\nLogin To View Results\n\nName\\*\n\nEmail ID\\*\n\nGrade\\*\n\nCity\\*\n\nView Result\n\n#### Comments\n### Leave a Comment [Cancel reply](https:\/\/byjus.com\/physics\/relation-between-electric-field-and-electric-potential\/#respond)\n\n![](https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2022\/12\/Vector-2219-2.png)\n\n#### Register with BYJU'S & Download Free PDFs\n#### Register with BYJU'S & Watch Live Videos\n![](https:\/\/cdn1.byjus.com\/byjusweb\/img\/landing-pages\/utils-icon\/play.svg)\n\n×","attrs_readable_markdown":"The electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – **“Electric field is the negative space derivative of electric potential.”**\n\n## Electric Field and Electric Potential\nThe relation between the electric field and [electric potential](https:\/\/byjus.com\/jee\/electric-potential-energy\/) is mathematically given by\n\nE \\= − d V d x\n\nWhere,\n\nE is the Electric field.\n\nV is the electric potential.\n\ndx is the path length.\n\n– sign indicates that the electric field is directed from higher potential to lower potential\n\n### Video Explanation:\n![](https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2021\/06\/Relation-between-Electric-Field-and-Potential.jpg)\n\n1,857\n\n### Direction of Electric Field\n- If the field is directed from lower potential to higher then the direction is taken to be positive.\n- If the field is directed from higher potential to lower potential then the direction is taken as negative.\n## Electric Field and Electric Potential Relation\n|   |   |   |\n|---|---|---|\n| **Test charge** | **Formula** | **Electric gradient** |\n| **Positive** | w q 0 \\= ∫ a b E → . d l → \\= V b − V a | Higher as you go closer towards test charge. |\n| **Negative** | w q 0 \\= ∫ a b E → . d l → \\= V a − V b | Higher as you go move away from test charge. |\n| **Equipotential surface** | w q 0 \\= ∫ a b E → . d l → \\= 0 | Electric potential is perpendicular to Electric field lines. |\n## Electric Field and Electric Potential Relation Derivation\nLet us study how to find the electric potential of the electric field is given. Work done in moving the test [charge](https:\/\/byjus.com\/physics\/electric-charge\/) q0 from a to b is given by-\n\nW ( q 0 a → b ) \\= ∫ a b F → . d l → \\= q 0 ∫ a b E → . d l →\n\nWhere,\n\n- F is the force applied\n- dl is the short element of the path while moving it from a to b.\n\n**The force can be written as charge times electric field**.\n\n\\= q 0 ∫ a b E → . d l →\n\nDividing both sides by test charge q0\n\nw q 0 \\= ∫ a b E → . d l →\n\nWork done by the test charge is the potential Va\\-Vb\n\n∫ a b E → . d l → \\= V a − V b\n\n**For equipotential surface, Va\\=Vb** thus,\n\n∫ a b E → . d l → \\= 0\n\nHope you understood the relation and conversion between Electric Field and Electric potential.\n\n**Physics Related Topics:**\n\n*Stay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.*","meta_canonical":null}

3. Robots.txt Check

Query:

Response:

4. Spam/Ban Check

Query:

Response:

5. Seen Status Check

ℹ️ Skipped - page is already crawled

📄

INDEXABLE

✅

CRAWLED

17 days ago

🤖

ROBOTS ALLOWED

Page Info Filters

Filter	Status	Condition	Details
HTTP status	PASS	`download_http_code = 200`	HTTP 200
Age cutoff	PASS	`download_stamp > now() - 6 MONTH`	0.6 months ago
History drop	PASS	`isNull(history_drop_reason)`	No drop reason
Spam/ban	PASS	`fh_dont_index != 1 AND ml_spam_score = 0`	ml_spam_score=0
Canonical	PASS	`meta_canonical IS NULL OR = '' OR = src_unparsed`	Not set

Page Details

Property	Value
URL	https://byjus.com/physics/relation-between-electric-field-and-electric-potential/
Last Crawled	2026-03-20 09:20:48 (17 days ago)
First Indexed	2019-03-20 04:39:35 (7 years ago)
HTTP Status Code	200
Meta Title	null
Meta Description	null
Meta Canonical	null
Boilerpipe Text	The electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.” Electric Field and Electric Potential The relation between the electric field and electric potential is mathematically given by E = − d V d x Where, E is the Electric field. V is the electric potential. dx is the path length. – sign indicates that the electric field is directed from higher potential to lower potential Video Explanation: 1,857 Direction of Electric Field If the field is directed from lower potential to higher then the direction is taken to be positive. If the field is directed from higher potential to lower potential then the direction is taken as negative. Electric Field and Electric Potential Relation Test charge Formula Electric gradient Positive w q 0 = ∫ a b E → . d l → = V b − V a Higher as you go closer towards test charge. Negative w q 0 = ∫ a b E → . d l → = V a − V b Higher as you go move away from test charge. Equipotential surface w q 0 = ∫ a b E → . d l → = 0 Electric potential is perpendicular to Electric field lines. Electric Field and Electric Potential Relation Derivation Let us study how to find the electric potential of the electric field is given. Work done in moving the test charge q 0 from a to b is given by- W ( q 0 a → b ) = ∫ a b F → . d l → = q 0 ∫ a b E → . d l → Where, F is the force applied dl is the short element of the path while moving it from a to b. The force can be written as charge times electric field . = q 0 ∫ a b E → . d l → Dividing both sides by test charge q 0 w q 0 = ∫ a b E → . d l → Work done by the test charge is the potential V a -V b ∫ a b E → . d l → = V a − V b For equipotential surface, V a =V b thus, ∫ a b E → . d l → = 0 Hope you understood the relation and conversion between Electric Field and Electric potential. Physics Related Topics: Stay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.
Markdown	![](https://www.facebook.com/tr?id=506170199519639&ev=PageView&noscript=1) The electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.” ## Electric Field and Electric Potential The relation between the electric field and [electric potential](https://byjus.com/jee/electric-potential-energy/) is mathematically given by \| \| \|---\| \| E \= − d V d x \| Where, E is the Electric field. V is the electric potential. dx is the path length. – sign indicates that the electric field is directed from higher potential to lower potential ### Video Explanation: ![](https://cdn1.byjus.com/wp-content/uploads/2021/06/Relation-between-Electric-Field-and-Potential.jpg) 1,857 ### Direction of Electric Field - If the field is directed from lower potential to higher then the direction is taken to be positive. - If the field is directed from higher potential to lower potential then the direction is taken as negative. ## Electric Field and Electric Potential Relation \| \| \| \| \|---\|---\|---\| \| Test charge \| Formula \| Electric gradient \| \| Positive \| w q 0 \= ∫ a b E → . d l → \= V b − V a \| Higher as you go closer towards test charge. \| \| Negative \| w q 0 \= ∫ a b E → . d l → \= V a − V b \| Higher as you go move away from test charge. \| \| Equipotential surface \| w q 0 \= ∫ a b E → . d l → \= 0 \| Electric potential is perpendicular to Electric field lines. \| ## Electric Field and Electric Potential Relation Derivation Let us study how to find the electric potential of the electric field is given. Work done in moving the test [charge](https://byjus.com/physics/electric-charge/) q0 from a to b is given by- W ( q 0 a → b ) \= ∫ a b F → . d l → \= q 0 ∫ a b E → . d l → Where, - F is the force applied - dl is the short element of the path while moving it from a to b. The force can be written as charge times electric field. \= q 0 ∫ a b E → . d l → Dividing both sides by test charge q0 w q 0 \= ∫ a b E → . d l → Work done by the test charge is the potential Va\-Vb ∫ a b E → . d l → \= V a − V b For equipotential surface, Va\=Vb thus, ∫ a b E → . d l → \= 0 Hope you understood the relation and conversion between Electric Field and Electric potential. Physics Related Topics: \| \| \|---\| \| [Electric Field Lines](https://byjus.com/physics/electric-field-lines/) \| \| [Electric Force](https://byjus.com/physics/electrical-force/) \| \| [Electrolytic Capacitor](https://byjus.com/physics/electrolytic-capacitor/) \| \| [Electric Circuit and Electrical Symbols](https://byjus.com/physics/electric-circuit-electrical-symbols/) \| Stay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist. Test Your Knowledge On Relation Between Electric Field And Electric Potential\! Q5 ![Quiz Image](https://cdn1.byjus.com/byjusweb/img/interactive-quiz/Quiz_cartoon.png) Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin\! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Start Quiz Congrats\! Visit BYJU’S for all Physics related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted View Quiz Answers and Analysis X Login To View Results Mobile Number\* Send OTP Did not receive OTP? Request OTP on Voice Call Login To View Results Name\* Email ID\* Grade\* City\* View Result #### Comments ### Leave a Comment [Cancel reply](https://byjus.com/physics/relation-between-electric-field-and-electric-potential/#respond) ![](https://cdn1.byjus.com/wp-content/uploads/2022/12/Vector-2219-2.png) #### Register with BYJU'S & Download Free PDFs #### Register with BYJU'S & Watch Live Videos ![](https://cdn1.byjus.com/byjusweb/img/landing-pages/utils-icon/play.svg) ×
Readable Markdown	The electric field exists if and only if there is an electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.” ## Electric Field and Electric Potential The relation between the electric field and [electric potential](https://byjus.com/jee/electric-potential-energy/) is mathematically given by E \= − d V d x Where, E is the Electric field. V is the electric potential. dx is the path length. – sign indicates that the electric field is directed from higher potential to lower potential ### Video Explanation: ![](https://cdn1.byjus.com/wp-content/uploads/2021/06/Relation-between-Electric-Field-and-Potential.jpg) 1,857 ### Direction of Electric Field - If the field is directed from lower potential to higher then the direction is taken to be positive. - If the field is directed from higher potential to lower potential then the direction is taken as negative. ## Electric Field and Electric Potential Relation \| \| \| \| \|---\|---\|---\| \| Test charge \| Formula \| Electric gradient \| \| Positive \| w q 0 \= ∫ a b E → . d l → \= V b − V a \| Higher as you go closer towards test charge. \| \| Negative \| w q 0 \= ∫ a b E → . d l → \= V a − V b \| Higher as you go move away from test charge. \| \| Equipotential surface \| w q 0 \= ∫ a b E → . d l → \= 0 \| Electric potential is perpendicular to Electric field lines. \| ## Electric Field and Electric Potential Relation Derivation Let us study how to find the electric potential of the electric field is given. Work done in moving the test [charge](https://byjus.com/physics/electric-charge/) q0 from a to b is given by- W ( q 0 a → b ) \= ∫ a b F → . d l → \= q 0 ∫ a b E → . d l → Where, - F is the force applied - dl is the short element of the path while moving it from a to b. The force can be written as charge times electric field. \= q 0 ∫ a b E → . d l → Dividing both sides by test charge q0 w q 0 \= ∫ a b E → . d l → Work done by the test charge is the potential Va\-Vb ∫ a b E → . d l → \= V a − V b For equipotential surface, Va\=Vb thus, ∫ a b E → . d l → \= 0 Hope you understood the relation and conversion between Electric Field and Electric potential. Physics Related Topics: Stay tuned with Byju’s for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.
Shard	185 (laksa)
Root Hash	6322170921519322385
Unparsed URL	com,byjus!/physics/relation-between-electric-field-and-electric-potential/ s443