A charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are [2023]

Question

A $10 μ C$ charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are [2023]

Smart Achievers · Accepted Answer

(2)

$Divide q = 10 μ C into two parts x and q - x .$

$F = \frac{K x (q - x)}{r^{2}}$

$For F to be maximum$

$\frac{d F}{d x} = \frac{K}{r^{2}} (q - 2 x) = 0 \Rightarrow x = \frac{q}{2}$

Solution

Q.
A $10 μ C$ charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are [2023]

1 $9 μ C, 1 μ C$

2 $5 μ C, 5 μ C$

3 $7 μ C, 3 μ C$

4 $8 μ C, 2 μ C$

Ans.
(2)

$Divide q = 10 μ C into two parts x and q - x .$

$F = \frac{K x (q - x)}{r^{2}}$

$For F to be maximum$

$\frac{d F}{d x} = \frac{K}{r^{2}} (q - 2 x) = 0 \Rightarrow x = \frac{q}{2}$

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Solution

Q. A 10 μC charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are [2023]

1 9 μC, 1 μC

2 5 μC, 5 μC

3 7 μC, 3 μC

4 8 μC, 2 μC

Ans. (2) Divide q=10 μC into two parts x and q-x. F=Kx(q-x)r2 For F to be maximum dFdx=Kr2(q-2x)=0⇒x=q2