Question

This is question ..plz help

Answer 1

Answer:

Given:

Charge Q is divided into 2 parts q and (Q-q) and separated by a distance d.

To find:

Distribution of charge such that the force of Repulsion is maximum.

Concept:

We shall find out the maximum force such that it's First - Order derivative will be equal to zero (Maxima will be reached).

Calculation:

Let k be Coulomb's constant,

$\bigstar \: \: F = \dfrac{kq(Q - q)}{ {r}^{2} }$

$= > \: \: F = \dfrac{k( {q}^{2} - Qq) }{ {r}^{2} }$

Now performing 1st order derivative with respect to q.

$= > \: \: \dfrac{dF}{dq} = \dfrac{k}{ {r}^{2} } \bigg \{ \dfrac{d ({q}^{2} - Qq ) }{dq} \bigg \}$

For Maxima , dF/dq = 0

$= > \: \: \dfrac{dF}{dq} = 2q - Q = 0$

$= > q = \dfrac{Q}{2}$

So maximum force of Repulsion only when

q = Q/2

Answer 2

Option B

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QUESTION :

A charge, Q is divided into two parts, "q" and " Q - q " and separated by a distance " R " The force of repulsion between them will be the maximum when :

1. q = Q / 4

2. q = Q / 2 √√√√√√√√

3. q = Q

4. q = Q / 8

SOLUTION :

First we need to find the electrostatic Force of repulsion between the two charges q and Q, separated by a distance , R

=> F = { 1 / 4 π e 0 } × { qQ - Q ^ 2 / R ^ 2 }

Now, we have the following condition..

The force of repulsion between them is maximum.

Thus :

$\dfrac{ \delta F }{ \delta Q } = 0$

Now, d/dx of K is 0

So, Q - 2q = 0

=> q = Q / 2