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

`W=-2.1405* 10^9\,J`

Step-by-step explanation:

Given:

electric field,
`E=148\,V.m^(-1)`

charge,
`Q=-13\,\mu C=-13* 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=√((x1-x2)^2+(y1-y2)^2)`

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

`d=√((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= (y2-y1)/(x2-x1)`

`\theta= tan^(-1)[(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* 279.63* cos\,47.75^(\circ)`

`\Delta V= 27826.06 V`

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

∴
`\Delta V=(W)/(Q)`

`W=(\Delta V)/(Q)`

Putting the respective values

`W=(27826.06 )/(-13* 10^(-6))`

`W=-2.1405* 10^9\,J`