JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2026
- Medium
- MCQ
Electric field in a region is given by \( \vec{E}=Ax\hat{i}+By\hat{j} \), where \( A=10~V/m^{2} \) and \( B=5~V/m^{2} \). If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is __________ V.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 2000
Step-by-step Solution
Detailed explanation
\( \vec{E}=10x\hat{i}+5y\hat{j} \)
\( V_{at(10,20)}=500~V \)
\( \Delta V=-\int\vec{E}\cdot d\vec{r} \)
\( 500-V_{0}=-\int_{(0,0)}^{(10,20)}(10x\hat{i}+5y\hat{j})\cdot(dx\hat{i}+dy\hat{j}) \)
\( 500-V_{0}=-[5x^{2}+\frac{5y^{2}}{2}]_{(0,0)}^{(10,20)} \)
\( V_{0}-500=(500+5\times\frac{400}{2})-(0-0) \)
\( V_{0}-500=500+1000 \)
\( V_{0}=2000V \)
\( V_{at(10,20)}=500~V \)
\( \Delta V=-\int\vec{E}\cdot d\vec{r} \)
\( 500-V_{0}=-\int_{(0,0)}^{(10,20)}(10x\hat{i}+5y\hat{j})\cdot(dx\hat{i}+dy\hat{j}) \)
\( 500-V_{0}=-[5x^{2}+\frac{5y^{2}}{2}]_{(0,0)}^{(10,20)} \)
\( V_{0}-500=(500+5\times\frac{400}{2})-(0-0) \)
\( V_{0}-500=500+1000 \)
\( V_{0}=2000V \)
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