Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  An electric field is given by kN/C. The potential of the point (1, –2), if

 An electric field is given byE _x = - 2 x^3 kN/C. The potential of the point (1, –2), if potential of the point (2, 4) is taken as zero, is                                                      

Option: 1

- 7.5 \times 10 ^3 V


Option: 2

7.5 \times 10 ^3 V


Option: 3

- 15 \times 10 ^3 V


Option: 4

15 \times 10 ^3 V


Answers (1)

best_answer

As we learned

Relation between E and V in integral form -

\dpi{100} dV=-\int_{r_{1}}^{r_{2}}\overrightarrow{E}\cdot \vec{d}r=-\int_{r_{1}}^{r_{2}}Edr\cos \theta

-

 

 

dV = - \bar E . d\bar r = - ( 2 x^3 \times 10 ^3 \hat i )(dx \hat i+ dy \hat j + dz \hat k ) = 2 x^3 \times 10 ^3 dx \\\\ \int_{0}^v{} dV = - \int_{2}^1{2x^3} \times 10 ^3 dx = V = 7.5 \times 10 ^3 V

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 paper 2 result out; Kerala’s Suryathejus, Gowri Sankar secure AIR 1 in BArch, BPlanning
anam-khan

JEE Mains 2026 session 2 registration ends tomorrow for BTech, BArch, BPlanning; exam from April 2 to 9

Latest Question