Question

2. The ratio of the amount of two charges is 2:3 and the distance between them is 5 cm. If the repulsive force acting between them is 96 dyne, then find the magnitude of each charge.​

Answer 1

Answer:

Q1= 13.334nC

Q2= 20nC

Explanation:

Here,

F = 96dyne

= 96÷100000 N [ 1N = 100000 dyne ]

= 0.00096 N

d = 5cm

= 0.05m [1m = 100cm]

Let, Q1= 2x

Q2= 3x

Now,

$f = c \times \frac{q1 \times q2}{ {d}^{2} } \\ = c \times \frac{2x \times 3x}{ {d}^{2} } \\ f \times \frac{ {d}^{2} }{2 \times 3 \times c} = {x}^{2} \\ {x}^{2} = 0.00096 \times \frac{ {0.05}^{2} }{6 \times 9 \times {10}^{9} } \\ = 4.44444 \times {10}^{ - 17} \\ x = \sqrt{4.44444 \times {10}^{ - 17} } \\ x = 6.6667 \times {10}^{ - 9}c \\ x = 6.6667nc$

Q1= 2x = 2× 6.6667nC

Q1= 2x = 2× 6.6667nC = 13.3334 nC

Q1= 2x = 2× 6.6667nC = 13.3334 nCand, Q2= 3x = 20nC

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