Question

find potential difference between points q and p when a uniform electric field of 100Vm inverse Act at 30 degree to x-axis given op is equal to 2 unit and oq is equal to 4 unit​

Answer 1

V= E×d ,we get

Vq-Vp=- Edpq

here, dpq = 2 cos 30°+2sin 30°

2×√3/2+4×1/2={√3+2)

(1.732+2)=3.732

Vq-Vp=-100×3.732

= -373.2V

Answer 2

Answer:

The potential difference between the points P and Q is 373,2V

Explanation:

Given a uniform electric field,  $E=100V/m$

Angle,  inverse Act at $\theta=30^{o}$

Length, $OP= 2 ~units$ and $OQ= 4 unit$

Let the distance between the surfaces be $d_{1}~\& ~d_{2$

The magnitude of electric potential between any two points is given by

$dV =E.dr$

$V_{q} -V_{p}= 100\times (OPcos\theta+OQsin\theta)$

$= 100\times (2cos30^{o} +4sin30^{o})$

$= 100\times (2\times \frac{\sqrt{3} }{2} +4\times \frac{1 }{2} )$

$= 100\times (\sqrt{3} +2)$

$\approx 373.2V$

Therefore, the potential difference between the points is 373,2V