Question

There exists a uniform electric field in the space as shown. Four points A, B, C and D are marked which are equidistant from the origin. If $V_A,V_B,V_C$ and $V_D$ are their potentials respectively, then

Answer 1

$hv=h{v}_0+k$

$\Rightarrow h\left(3\times {10}^{14}\right)=h\left(v_0\right)+2k$

and $h\left(2\times {10}^{14}\right)=h\left(v_0\right)+k$

$\Rightarrow \frac{3\times {10}^{14}-v_0}{2\times {10}^{14}-v_0}=2$

$\Rightarrow 3\times {10}^{14}-v_0=2\times {10}^{14}-v_0$

$\Rightarrow v_0=1\times {10}^{14}$

We get, $v_0=1\times {10}^{14}$