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Question: Two small conducting spheres of equal radius have charges \[ + 10\mu C\] and \[ - 20\mu C\] respecti...

Two small conducting spheres of equal radius have charges +10μC + 10\mu C and −20μC - 20\mu C respectively and placed at a distance of RR from each other and experience force F1{F_1} . If they were brought in contact and separated to the same distance, the experience forced F2{F_2} . The ratio of F1 to F2{F_1}{\text{ }}to{\text{ }}{F_2} is:
A. 1:81:8
B. −8:1 - 8:1
C. 1:21:2
D. −2:1 - 2:1

Explanation

Solution

As we can see that in question, two charges are placed and also there are two forces experiencing. Two forces experienced are: when charges are not in contact then force F1{F_1} occurs and when charges are in contact then force F2{F_2} occurs. So, we will apply the formula of force in the terms of given charges separately for both cases. And then we can find the ratio of both the forces.

Complete step by step answer:
According to the question, the first charge on the sphere, q1=+10μC{q_1} = + 10\mu C.
Second charge on the sphere, q2=−20μC{q_2} = - 20\mu C
Distance between charges, d=Rd = R
So, we know that, the formula of force in terms of given charges:
F1=kq1q2R2{F_1} = \dfrac{{k{q_1}{q_2}}}{{{R^2}}}
⇒F1=9×109×10×10−6×(−20)×10−6R2\Rightarrow {F_1} = \dfrac{{9 \times {{10}^9} \times 10 \times {{10}^{ - 6}} \times ( - 20) \times {{10}^{ - 6}}}}{{{R^2}}}

Now, if they are brought in contact and separated to some distance they experience Force F2{F_2} :
When the charge will be on some spheres after contact:-
Q=q1+q2 ⇒Q=−10×10−62 ⇒Q=−5×10−6c Q = {q_1} + {q_2} \\\ \Rightarrow Q = \dfrac{{ - 10 \times {{10}^{ - 6}}}}{2} \\\ \Rightarrow Q = - 5 \times {10^{ - 6}}c \\\
So, the new force will be:-
F2=kQ2d2 ⇒F2=9×109×(5×10−6)2R2 {F_2} = \dfrac{{k{Q^2}}}{{{d^2}}} \\\ \Rightarrow {F_2} = \dfrac{{9 \times {{10}^9} \times {{(5 \times {{10}^{ - 6}})}^2}}}{{{R^2}}} \\\
Now, we have both F1 and F2{F_1}\,and\,{F_2} , so we can find the ratio of F1 to F2{F_1}{\text{ }}to{\text{ }}{F_2} -
F1F2=9×109×10×10−6×(−20)×10−6R29×109×(5×10−6)2R2 ⇒F1F2=−10×2025 ∴F1F2=−81\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{\dfrac{{9 \times {{10}^9} \times 10 \times {{10}^{ - 6}} \times ( - 20) \times {{10}^{ - 6}}}}{{{R^2}}}}}{{\dfrac{{9 \times {{10}^9} \times {{(5 \times {{10}^{ - 6}})}^2}}}{{{R^2}}}}} \\\ \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{ - 10 \times 20}}{{25}} \\\ \therefore \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{ - 8}}{1}
Now, we can write the upper equation in the form of ratio or we can say that the ratio of F1 to F2{F_1}{\text{ }}to{\text{ }}{F_2} is −8:1 - 8:1 .

Hence, the correct option is B.

Note: The force acting on a charge is proportional to its size and inversely proportional to the square of the distance between the two charges. The vector sum of all individual forces acting on the charges is the force acting on a point charge due to many charges.