Question

Four charges +4uC, -2uC, -4uC and +2uC are placed at the corner A,B,C,and D respectively of a square of side 20 cm . Calculate :

(a) Net force on +2uC charge.
(b) Net force on a fifth positive charge of +5uC placed at the center of the square.

Answer 1

Answer:

Part a)

Net force on +2uC charge is 2.58 N

Part b)

Net force on a fifth positive charge of +5uC placed at the center of the square is 20.12 N

Explanation:

As we know that force between two charges at distance "d" is given as

$F = \frac{kq_1q_2}{d^2}$

here force due to corner A is given as

$F_1 = \frac{(9\times 10^9)(4\mu C)(2\mu C)}{0.20^2}$

$F_1 = 1.8 N \hat i$

Force due to corner B is given as

$F_2 = \frac{(9\times 10^9)(2\mu C)(2\mu C)}{2(0.20)^2}$

$F_2 = 0.45(\frac{1}{\sqrt2}(-\hat i - \hat j))$

Force due to corner C is given as

$F_3 = \frac{(9\times 10^9)(4\mu C)(2\mu C)}{0.20^2}$

$F_3 = 1.8 N (-\hat j)$

now we have

$F_{net} = F_1 + F_2 + F_3$

$F_{net} = 1.8 \hat i + (-0.32 \hat i - 0.32 \hat j) - 1.8 \hat j$

$F_{net} = 1.48 \hat i - 2.12\hat j$

$F_{net} = \sqrt{1.48^2 + 2.12^2}$

$F_{net} = 2.58 N$

Part b)

If another charge is placed at the center of the square then net force on it due to corner A and C

$F_1 = 2(\frac{(9\times 10^9)(4\mu C)(5\mu C)}{0.5(0.20)^2})$

$F_1 = 18 N$

Now for net force due to corner C and D is given as

$F_2 = 2(\frac{(9\times 10^9)(2\mu C)(5\mu C)}{0.5(0.20)^2})$

$F_2 = 9 N$

Now resultant of F1 and F2 is given as

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{9^2 + 18^2}$

$F = 20.12 N$

#Learn

Topic : electrostatic force

https://brainly.in/question/13169785

Answer 2

Answer:

Hope it helps

Explanation: