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Question: The capacity of an isolated sphere is increased \(n\) times when it is enclosed by an earthed concen...

The capacity of an isolated sphere is increased nn times when it is enclosed by an earthed concentric sphere. The ratio of their radii is
(A) nn−1\dfrac{n}{n-1}
(B) n2n−1\dfrac{{{n}^{2}}}{n-1}
(C) 2nn+1\dfrac{2n}{n+1}
(D) 2n+1n+1\dfrac{2n+1}{n+1}

Explanation

Solution

Hint : To solve this question we have to find the capacitance of two spheres both is concentric. After finding capacitance we have compared both of the equations to find the ratio of their radius.

**Complete step by step answer: ** Let the radius of a first sphere is R1{{R}_{1}} and radius of other sphere is R2{{R}_{2}}

Here is the potential on the surface of a sphere is V=KQR1V=\dfrac{KQ}{{{R}_{1}}}
We let the second plate on the infinite and potential so V1=0{{V}_{1}}=0
Now we calculate the potential difference between the surface of the sphere and plate on the infinite. Let the difference between both refer to ΔV\Delta V.
So we can write the equation as
ΔV=V−V1\Delta V=V-{{V}_{1}}
Here we put the values in the equation
ΔV=KQR1−0\Delta V=\dfrac{KQ}{{{R}_{1}}}-0
Then we get,
ΔV=KQR1\Delta V=\dfrac{KQ}{{{R}_{1}}}
Now we put the value ofKKin the above equation
K=14πε0K=\dfrac{1}{4\pi {{\varepsilon }_{0}}}
After putting the value of K we get,
ΔV=14πε0R1⋅Q\Delta V=\dfrac{1}{4\pi {{\varepsilon }_{0}}{{R}_{1}}}\cdot Q

Now we calculate charge in the sphere using the following equation
Q=CΔVQ=C\Delta V
Here we put the value of the potential difference in the above equation
Q=C1⋅Q4πε0R1Q={{C}_{1}}\cdot \dfrac{Q}{4\pi {{\varepsilon }_{0}}{{R}_{1}}}
After solving this equation we get
C1=4πε0R1{{C}_{1}}=4\pi {{\varepsilon }_{0}}{{R}_{1}}
Here we get the capacitance in the first sphere.
Now we calculate capacitance on the outer sphere.

The charge on the outer surface of the sphere is negative. And in the inner sphere charge is positive.
So we calculate total potential by both spheres,
V=−KQR2V=\dfrac{-KQ}{{{R}_{2}}}
V1=KQR2{{V}_{1}}=\dfrac{KQ}{{{R}_{2}}}
So the total potential of both sphere is
ΔV=−KQR2+KQR1\Delta V=\dfrac{-KQ}{{{R}_{2}}}+\dfrac{KQ}{{{R}_{1}}}
Now take K as a common value in the equation
ΔV=K(1R1−1R2)\Delta V=K\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)
Now we put the value of KK in the above equation,
ΔV=Q4πε0(1R1−1R2)\Delta V=\dfrac{Q}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)
Here we calculate charge in the sphere,
Q=C2ΔVQ={{C}_{2}}\Delta V
Here we put the value of the total potential in both sphere,
Q=C2⋅Q4πε0(R2−R1R1R2)Q={{C}_{2}}\cdot \dfrac{Q}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{R}_{2}}} \right)
After solving this equation we get the capacitance of both spheres,
C2=4πε0R1R2R2−R1{{C}_{2}}=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}}
As mentioned in the question the value capacitance of C2{{C}_{2}} change in nn times of C1{{C}_{1}}
nC1=n⋅4πε0R1n{{C}_{1}}=n\cdot 4\pi {{\varepsilon }_{0}}{{R}_{1}}
Now comparing both equations,
C2=4πε0R1R2R2−R1=nC1=n⋅4πε0R1{{C}_{2}}=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}}=n{{C}_{1}}=n\cdot 4\pi {{\varepsilon }_{0}}{{R}_{1}}
After solving this equation we get,
⇒R2R2−R1=n\Rightarrow \dfrac{{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}}=n
After solving this we get,
⇒R2=nR2−nR1\Rightarrow {{R}_{2}}=n{{R}_{2}}-n{{R}_{1}}
Take R2{{R}_{2}} in the other side so we can take R2{{R}_{2}} it as a common term
nR1=nR2−R2n{{R}_{1}}=n{{R}_{2}}-{{R}_{2}}
After taking R2{{R}_{2}} we get
nR1=(n−1)R2n{{R}_{1}}=(n-1){{R}_{2}}
After further solving
R2R1=nn−1\dfrac{{{R}_{2}}}{{{R}_{1}}}=\dfrac{n}{n-1}
Here we get the ratio between both spheres is =nn−1=\dfrac{n}{n-1}

Note: To solve this type of question we have to find the potential difference of both spheres. After that, we calculate charge in both spheres. The charge of the outer sphere is negative but the charge of the inner sphere is positive. Then we calculate the capacitance of both spheres then compares both of them to calculate the ratio of the radius. If we follow these steps we can easily solve this question.