Answer:
[tex]W=-2.1405\times 10^9\,J[/tex]
Explanation:
Given:
electric field, [tex]E=148\,V.m^{-1}[/tex]
charge, [tex]Q=-13\,\mu C=-13\times 10^{-6}\,C[/tex]
initial position coordinates, [tex]p1 =(-18,-131)[/tex]
final position coordinates, [tex]p2 =(107,76)[/tex]
We find the distance through which the charge has been moved:
[tex]d=\sqrt{(x1-x2)^2+(y1-y2)^2}[/tex]
Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.
[tex]d=\sqrt{(107-(-81))^2+(76-(-131))^2}[/tex]
[tex]d= 279.63\,m[/tex]
Now we need the angle through which displacement is made with respect to the direction of electric field.
[tex]tan\,\theta= \frac{y2-y1}{x2-x1}[/tex]
[tex]\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ][/tex]
[tex]\theta= 47.75^{\circ}[/tex]
Now from the relation between the change in potential difference:
[tex]\Delta V= E.d.cos\,\theta[/tex]
[tex]\Delta V= 148\times 279.63\times cos\,47.75^{\circ}[/tex]
[tex]\Delta V= 27826.06 V [/tex]
∵The change in voltage is defined as the work done per unit charge.
∴[tex]\Delta V=\frac{W}{Q}[/tex]
[tex]W=\frac{\Delta V}{Q}[/tex]
Putting the respective values
[tex]W=\frac{27826.06 }{-13\times 10^{-6}}[/tex]
[tex]W=-2.1405\times 10^9\,J[/tex]
