Question

A charge of 2×10^-7C is acted upon by force of 0.1N . Determine the distance to the other charge of 4.5×10^-7 C both the charges are in vacuum

Answer 1

Answer:

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Explanation:

Answer: d

Explanation: F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) =

We get r = 0.09m.

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Answer 2

A charge of 2×10^-7C is acted upon by force of 0.1N and the other charge is 4.5×10^-7 C.

We have to find the distance between both the charges when they are placed in vacuum.

Given; q1 = 2×10^-7 C, q2 = 4.5×10^-7 C and F = 0.1N

F = k (q1q2)/r²

F = 1/4πεo q1q2/r²

Substitute the values,

0.1 = (9 × 10^9 × 2 × 10^-7 × 4.5 × 10^-7)/r²

r² = (81 × 10^(-14+9))/0.1

r² = (81 × 10^-5)/0.1

r² = 81 × 10^-4

Square root both sides,

(√r)² = √(81 × 10^-4)

r = 9 × 10^-2

r = 0.09

Therefore, the distance between both the charges is 0.09 m.

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Coulomb's Law:

According to it magnitude of force between two static point charges is -

• directly proportional to product of magnitude of charges i.e.

F ∝ q1q2

• inversely proportional to square of distance between two charges i.e.

F ∝ 1/r²

So,

F ∝ q1q2/r²

F = k q1q2/r²

Here,

k = 1/4πεo = 9 × 10^9 Nm²C-²

εo = 8.854 × 10^-12 C²N-¹m-²