Question

ABCD is a square of side 0.2m. Charges 2nC, 4nC, 6nC and 1nC are placed at the corners A, B, C and D respectively. Calculate the magnitude of force on 1nC charge.​

Answer 1

ABC bhghghghhbabc efg hijk

Answer 2

Given:

Side of square = 0.2m

Charges on 4 corners are 2nC, 4nC , 6nC and 1nC

To find:

Force on 1nC charge

Solution:

The charges are placed at 4 corners as shown in the figure attached.

Consider charge 2nC on A , 4nC on B  6nC on C and 1nC on D.

Force ( $F_C_D$   )acting on D , due to charge at C (6nC) will be :

$F_C_D = \frac{Fq_1q_2}{r^2}$ , where F is Coloumb's constant.

$F_C_D = \frac{9\times 10^9 \times 1\times 6 \times 10^-^1^8}{0.2^2}$   $= 1.348\times10^-^6 N$

Similarly force on D (1nC) due to to charge at A ( 2nC) will be:

$F_A_D = \frac{9\times 10^9 \times 1\times 2 \times 10^-^1^8}{0.2^2}$  = $0.44\times10^-^6 N$

and force on D (1nC) due to to charge at B ( 4nC) will be:

$F_B_D = \frac{9\times 10^9 \times 1\times 4 \times 10^-^1^8}{0.2^2}$ = $0.898\times10^-^6 N$

Now components of forces acting on D due to B will be $F_bsin45$ and $F_Bcos45$ both in perpendicular direction at D

$\therefore$ force on D in vertical direction :

$F_1 = F_A_D + 0.898\times10^-^6 \sin45$ $= 0.44\times10^-^6 + 0.898\times10^-^6\times 0.707$

                                                $= 1.07 \times 10^-^6 N$

and

$F_2 = F_C_D + 0.898\times10^-^6 \cos45$ $=1.348 \times10^-^6 + 0.898\times10^-^6 \times0.707$

                                                $= 1.982 \times 10^-^6 N$

$\therefore$ Net force on D, $F = \sqrt{F_1^{2} + {F_2^{2}}$

                                  = $2.25 \times 10^-^6 N$