Question

In an electric field $\overrightarrow{E}=(2\hat{i}+4\hat{j})N/C$, electric potential at origin is 0 V. Match the following two columns :

Answer 1

The electric potential between two points a and b is
$V_b-V_a=-{\int }_a^b\overrightarrow{E}.\overrightarrow{dr}$
A) $V_{(4,0)}-V_{(0,0)}=-{\int }_{(0,0)}^{(4,0)}(2\hat{i}+4\hat{j}).(dx\hat{i}+dy\hat{j})=-2{\int }_0^{4}2dx$ (as y is constant so $dy=0$)
or $V_{(4,0)}-0=-8\Rightarrow V_{(4,0)}=-8V$
Similarly for
B) $V_{(-4,0)}-V_{(0,0)}=-2{\int }_0^{-4}2dx$
or $V_{(-4,0)}-0=8\Rightarrow V_{(-4,0)}=8V$
C)$V_{(0,4)}-V_{(0,0)}=-{\int }_{(0,0)}^{(0,4)}(2\hat{i}+4\hat{j}).(dx\hat{i}+dy\hat{j})=-4{\int }_0^{4}2dy$ (as x is constant so $dx=0$)
or $V_{(0,4)}-0=-16\Rightarrow V_{(0,4)}=-16V$
Similarly for
B) $V_{(0,-4)}-V_{(0,0)}=-4{\int }_0^{-4}2dx$
or $V_{(0,-4)}-0=16\Rightarrow V_{(-4,0)}=16V$
Ans: $A\to 2,B\to 1,C\to 4,D\to 3$