Question

Two small identical metallic balls having charges \(q\) and \(-2q\) are kept far apart at a separation \(r\). They are brought in contact and then separated at a distance \(\frac{r}{2}\). Compared to the initial force \(F\), they will now:

• A. attract with a force \(\frac{F}{2}\)
• B. repel with a force \(\frac{F}{2}\)
• C. repel with a force \(F\)
• D. attract with a force \(F\)

Hint:

For identical conducting spheres brought into contact, always conserve total charge and divide it equally between the spheres. After finding the new charges, use Coulomb's law again with the new separation and compare the forces carefully.

Answer 1

The Correct Option is B

Concept: When two identical conducting spheres are brought into contact, the total charge is redistributed equally between them because both spheres have the same capacitance. The final charge on each sphere is given by \[ q_f=\frac{\text{Total Charge}}{2}. \] The electrostatic force between two point charges is determined by Coulomb's law: \[ F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2}. \] The nature of the force depends upon the signs of the charges:
• Like charges repel each other.
• Unlike charges attract each other. Therefore, we first calculate the initial force, then determine the new charges after contact, and finally compare the new force with the original force.

Step 1: Calculate the initial electrostatic force between the two spheres. Initially, the charges on the spheres are \[ q_1=q \] and \[ q_2=-2q. \] The magnitude of the initial force is \[ F=\frac{1}{4\pi\varepsilon_0} \frac{|q(-2q)|}{r^2}. \] Hence, \[ F=\frac{1}{4\pi\varepsilon_0} \frac{2q^2}{r^2}. \] Since the charges are of opposite signs, the force is attractive.

Step 2: Determine the charge on each sphere after contact. The total charge of the system is \[ q+(-2q)=-q. \] Since the spheres are identical, this total charge gets equally shared. Therefore, the charge on each sphere after contact becomes \[ q_f=\frac{-q}{2}. \] Thus, after separation, \[ q_1'=-\frac{q}{2}, \qquad q_2'=-\frac{q}{2}. \]

Step 3: Calculate the new force when the spheres are separated by \(\frac{r}{2}\). The new separation is \[ r'=\frac{r}{2}. \] Applying Coulomb's law, \[ F' = \frac{1}{4\pi\varepsilon_0} \frac{\left(-\frac{q}{2}\right)\left(-\frac{q}{2}\right)} {\left(\frac{r}{2}\right)^2}. \] Since both charges are negative, their product is positive: \[ F' = \frac{1}{4\pi\varepsilon_0} \frac{\frac{q^2}{4}} {\frac{r^2}{4}}. \] The factor \(\frac{1}{4}\) cancels: \[ F' = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}. \]

Step 4: Compare the new force with the original force. We have \[ F= \frac{1}{4\pi\varepsilon_0} \frac{2q^2}{r^2} \] and \[ F'= \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}. \] Therefore, \[ \frac{F'}{F} = \frac{\frac{q^2}{r^2}} {\frac{2q^2}{r^2}} = \frac{1}{2}. \] Hence, \[ F'=\frac{F}{2}. \] Since both spheres carry negative charges after contact, the force is repulsive. Therefore, the spheres repel each other with a force \[ \boxed{\frac{F}{2}}. \] Hence, the correct answer is \[ \boxed{\text{(B) repel with a force } \frac{F}{2}}. \]