Question

A, B, C, D, P and Q are points in a uniform electric field. Thee potentials at these points are $V\left(A\right)=2V.V\left(P\right)=V\left(B\right)=V\left(D\right)=5V.V\left(C\right)=8V.$ . The electric field at P is

• A. $10V{m}^{-1}alongPQ$
• B. $15\sqrt{2}V{m}^{-1}alongPA$
• C. $5V{m}^{-1}alongPC$
• D. $5V{m}^{-1}alongPA$

Answer 1

The correct option is B $15\sqrt{2}V{m}^{-1}alongPA$

$V\left(A\right)=2V$

$V\left(P\right)=V\left(B\right)=V\left(d\right)=5V$

$V\left(C\right)=8V$

There will be no electric field along $BPD$ because all the are at same potential so, potential difference is zero

The electric field will only be along $APC$

Potential difference along $APC=V\left(C\right)-V\left(A\right)$

$=8-2$

$=6V$

Distance $AC=0.2\sqrt{2}m$

$\therefore$ Electric field along $APC=\frac{△V}{AC}$

$=\frac{60}{2\sqrt{2}}=\frac{30}{\sqrt{2}}$

$[E=15\sqrt{2}V/m]$

Hence option (B) is correct