Question

1. Two small conducting spheres of equal radius have charges +10μC and -20μC respectively and placed at a distance R from each other experience force F1. If they are brought in contact and separated to the same distance, they experience force F2. The ratio of F1 to F2 is:
(a) 1 : 8
(b) -8 : 1
(c) 1 : 2
(d) -2 : 1

Answer 1

f 1 = 9×10^9 200×10^-6%R^2 and F 2 = -9×10^9 × 10^-6% R^2 now divide F1 and F2 then you will get 4 the option ; CONCEPT behind it is when two charges are attached then they will get same charge as -20+10= -10 C

Answer 2

Answer:

(b). The ratio of the F₁ and F₂ is -8:1.

Explanation:

Given that,

First charge $q_{1}=+10\mu C$

Second charge $q_{2}=-20\mu C$

Distance d= R

Two small conducting spheres of equal radius have charges +10μC and -20μC respectively and placed at a distance R from each other experience force $F_{1}$

The force is

$F_{1}=\dfrac{kq_{1`}q_{2}}{d^2}$

$F_{1}=\dfrac{9\times10^{9}\times10\times10^{-6}\times(-20)\times10^{-6}}{R^2}$

If they are brought in contact and separated to the same distance, they experience force $F_{2}$

When the charge will be same on the spheres after contact.

The charge will be

$Q=\dgfrac{q_{1}+q_{2}}{2}$

$Q=\dfrac{10\times10^{-6}}{2}$

$Q=-5\times10^{-6}\ C$

The new force will be

$F_{2}= \dfrac{kQ^2}{d^2}$

$F_{2}=\dfrac{9\times10^{9}\times(5\times10^{-6})^2}{r^2}$

The ratio of the F₁ and F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{\dfrac{9\times10^{9}\times10\times10^{-6}\times(-20)\times10^{-6}}{R^2}}{\dfrac{9\times10^{9}\times(5\times10^{-6})^2}{R^2}}$

$\dfrac{F_{1}}{F_{2}}=\dfrac{-10\times20}{25}$

$\dfrac{F_{1}}{F_{2}}=\dfrac{-8}{1}$

Hence, The ratio of the F₁ and F₂ is -8:1.