Question

A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move the charge Q =-13 μC from x1 = -81 m, y1 = -131 m, to x2 = 107 m, y2 = 76 m? (Please note: the charge is given in micro-Coulomb. The symbol in front of the C should be a Greek letter mu.)

Answer 1

Answer:

$W=-2.1405\times 10^9\,J$

Explanation:

Given:

electric field, $E=148\,V.m^{-1}$

charge, $Q=-13\,\mu C=-13\times 10^{-6}\,C$

initial position coordinates, $p1 =(-18,-131)$

final position coordinates,   $p2 =(107,76)$

We find the distance through which the charge has been moved:

$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

$d=\sqrt{(107-(-81))^2+(76-(-131))^2}$

$d= 279.63\,m$

Now we need the angle through which displacement is made with respect to the direction of electric field.

$tan\,\theta= \frac{y2-y1}{x2-x1}$

$\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]$

$\theta= 47.75^{\circ}$

Now from the relation between the change in potential difference:

$\Delta V= E.d.cos\,\theta$

$\Delta V= 148\times 279.63\times cos\,47.75^{\circ}$

$\Delta V= 27826.06 V$

∵The change in voltage is defined as the work done per unit charge.

∴$\Delta V=\frac{W}{Q}$

$W=\frac{\Delta V}{Q}$

Putting the respective values

$W=\frac{27826.06 }{-13\times 10^{-6}}$

$W=-2.1405\times 10^9\,J$