Question

Two point charges of charge values Q and q are placed at distance x and x/2 respectively from a third charge of charge value 4q, all charges being in the same straight line. Calculate the magnitude and nature of charge Q such that the net force experienced by the charge q is zero. [Ans Q = q/4 , +ive ]

Answer 1

Pardon my handwriting XP

Answer 2

Answer:

The magnitude of the charge is "4q" and the nature is "negative".

Explanation:

The force acting between the charges is given as,

$F=\frac{kQ_1Q_2}{r^2}$     (1)

Where,

F= force acting between the charges

k=coulombs constant

Q₁, Q₂=respective charges

r=distance between the charges Q₁ and Q₂

Now the force acting between the charges "q" and "4q" is given as,

$F_1=\frac{k\times q\times 4q}{(\frac{x}{2} )^2}$

$F_1=\frac{16kq^2}{x^2}$      (2)

The force acting between the charges "q" and "Q" is given as,

$F_2=\frac{k\times q\times Q}{(\frac{x}{2} )^2}$

$F_2=\frac{4kQq}{x^2}$      (3)

In the question, it is given that the net force is zero. So,

$F_1+F_2=0$    (4)

By substituting the required equations in equation (4) we get;

$\frac{16kq^2}{x^2}+\frac{4kQq}{x^2}=0$

$\frac{4kQq}{x^2}=-\frac{16kq^2}{x^2}$

$Q=-4q$

Hence, the magnitude of the charge is "4q" and the nature is "negative".

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