Question

Calculate the electric field (magnitude and direction) at the upper right corner of a square 1.22 m on a side if the other three corners are occupied by 2.40×10−6 C charges. Assume that the positive x-axis is directed to the right.

Answer 1

Step-by-step explanation:

It is mentioned that the magnitude of given charge is
`2.40 * 10^(-6) C` and side of the square is 1.22 m.

Hence, we will calculate the electric field due to charge at point A as follows.

`E_(a) = k(q)/(r^(2))`

=
`(9 * 10^(9) * 2.4 * 10^(-6))/((1.22)^(2))`

=
`14.512 * 10^(3) N/C` (acts along y-axis)

Electric field at D due to the charge at C is as follows.

`E_(c) = k(q)/(r^(2))`

=
`(9 * 10^(9) * 2.4 * 10^(-6))/((1.22)^(2))`

=
`14.512 * 10^(3) N/C` (acts along y-axis)

Electric field at D due to the charge at B is as follows.

`E_(d) = k(q)/(r^(2))`

=
`(9 * 10^(9) * 2.4 * 10^(-6))/((1.7253)^(2))`

=
`7.2564 * 10^(3) N/C`

Now, we will resolve it into two components.

`E_(bx) = E_(b) Cos 45 = 5.1318 * 10^(3) N/C`

`E_(by) = E_(b) Sin 45 = 5.1318 * 10^(3) N/C`

Now, the resultant x component is as follows.

`E_(x) = (14.512 + 5.1318) * 10^(3) = 19.6438 * 10^(3) N/C`

Resultant y component is given as follows.

`E_(y) = (14.512 + 5.1318) * 10^(3) = 19.6438 * 10^(3) N/C`

Therefore, resultant electric field at point D is as follows.

`E = \sqrt{E^(2)_(x) + E^(2)_(y)}`

=
`\sqrt{(19.6438 * 10^(3))^(2) + (19.6438 * 10^(3))^(2)}`

=
`27.780 * 10^(3) N/C`

And, the direction of electric field is

`\theta = tan^(-1) ((E_(y))/(E_(x)))`

=
`tan^(-1) (1)`

=
`45^(o)`

This means that net electric field makes an angle of
`135^(o)` with positive x-axis in counter clockwise direction.