Question

My Notes Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of q/Q will the electrostatic force between the two spheres be maximized?

Answer 1

Answer:

the electric force between the spheres is maximum when,

$\rm \dfrac qQ=\dfrac 12.$

Explanation:

Initally, the charge on first sphere = Q.

Now, a portion of q charge is transferred to the second sphere,therefore,

• The charge acquired by the second sphere, $\rm q_2$ = q.

• The charge remained on the first sphere, $\rm q_1$ = Q-q.

Let the two spheres are separated by distance $\rm a$.

According to the Coulomb's law, the electrostatic force of interaction between the two static point charges $\rm q_1$ and $\rm q_2$, separated by a distance $\rm r$ is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where,

k is the Coulomb's constant.

It is given that both the spheres can be treated as particles and are fixed with a certain separation.

Therefore, the electrostatic force of interaction between the two spheres is given as:

$\rm F = \dfrac{k(Q-q)q}{a^2}=\dfrac{k(Qq-q^2)}{a^2.}$

The electrostatic force between the two spheres is extremum for the value of $\rm q=q_o$, when $\rm\left ( \dfrac{dF}{dq}\right )_{q=q_o}=0.$

$\rm \left ( \dfrac{dF}{dq}\right )_{q=q_o}=\left [ \dfrac{d}{dq}\left (\dfrac{k(Qq-q^2)}{a^2} \right ) \right ]_{q=q_o}\\\\=\left [ \dfrac{k}{a^2}\dfrac{d}{dq}\left (Qq-q^2 \right ) \right ]_{q=q_o}\\=\dfrac{k}{a^2}(Q-2q_o).$

For,  $\rm\left ( \dfrac{dF}{dq}\right )_{q=q_o}=0,$

$\rm \dfrac{k}{a^2}(Q-2q_o)=0\\\\\Rightarrow Q-2q_o=0\\q_o=\dfrac Q2.$

The electrostatic force is maximum when,

$\rm\left ( \dfrac{d^2F}{dq^2}\right )_{q=q_o}<0.$

$\rm\left ( \dfrac{d^2F}{dq^2}\right )_{q=q_o}=\left ( \dfrac{d}{dq}\left (\dfrac{dF}{dq}\right )\right )_{q=q_o}\\\\=\left ( \dfrac{d}{dq}\left (\dfrac{k}{a^2}(Q-2q)\right )_{q=q_o}\\\\=\dfrac{k}{a^2}(-2).$

$\rm \text{Since, k and a are positive constants, therefore, }\\\left ( \dfrac{d^2F}{dq^2}\right )_{q=q_o}<0$

Thus, the electric force between the spheres is maximum when,

$\rm q=q_o = \dfrac Q2\\\\ i.e.,\ \dfrac qQ=\dfrac 12.$

Answer 2

Answer:

$\dfrac{q}{Q}=0.5$ m

Explanation:

Given:

Initial charge on the tiny sphere=Q

Charge transferred from tiny sphere to another sphere=q

According to coulombs law the electrostatic force between the two spheres is given by

$F=\dfrac{k(Q-q)q}{r^2}$

Where r is the fixed distance between them.

For getting maximum force we will differentiate the force with respect to q

$F=\dfrac{d(\dfrac{k(Q-q)q}{r^2})}{dq}=0\\ \dfrac{q}{Q}=\dfrac{1}{2}\\ \dfrac{q}{Q]=0.5$

Hence we calculated the the ratio for maximum force between them