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Question: Two metallic spheres of radii \[1\,{\text{cm}}\] and \[3\,{\text{cm}}\] are given charges of \[ - 1 ...

Two metallic spheres of radii 1 cm1\,{\text{cm}} and 3 cm3\,{\text{cm}} are given charges of −1×10−2 C - 1 \times {10^{ - 2}}\,{\text{C}} and 5×10−2 C5 \times {10^{ - 2}}\,{\text{C}}, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is-
A. 3×10−2 C3 \times {10^{ - 2}}\,{\text{C}}
B. 4×10−2 C4 \times {10^{ - 2}}\,{\text{C}}
C. 1×10−2 C1 \times {10^{ - 2}}\,{\text{C}}
D. 2×10−2 C2 \times {10^{ - 2}}\,{\text{C}}

Explanation

Solution

Use the formula for the potential on the metallic sphere. This formula gives the relation between the potential, charge and radius of the metallic sphere. When the two metallic spheres are connected by a conducting wire, they exchange charges until the potential on both of them becomes equal. Hence, the potential of two spheres after connecting the wires is the same.

Formula used:
The potential VV on the sphere is given by
V=kqrV = \dfrac{{kq}}{r}
Here, kk is the constant, qq is the charge and rr is the radius of the sphere.

Complete step by step answer:
We have given that the metallic sphere of radius 1 cm1\,{\text{cm}} has charge −1×10−2 C - 1 \times {10^{ - 2}}\,{\text{C}}.
r1=1 cm{r_1} = 1\,{\text{cm}}
q1=−1×10−2 C{q_1} = - 1 \times {10^{ - 2}}\,{\text{C}}
We have given that the metallic sphere of radius 3 cm3\,{\text{cm}} has charge 5×10−2 C5 \times {10^{ - 2}}\,{\text{C}}.
r2=3 cm{r_2} = 3\,{\text{cm}}
q2=5×10−2 C{q_2} = 5 \times {10^{ - 2}}\,{\text{C}}
When the two metallic spheres are connected with a conducting wire, they exchange charge until the potential on both of them becomes equal.
V1=V2{V_1} = {V_2} …… (2)
Here, V1{V_1} and V2{V_2} are the potentials on the first and second metallic sphere.
When the two spheres are connected, they exchange charges and their final charges is the sum of the two charges on the metallic spheres.
q=q1+q2q = {q_1} + {q_2}
Substitute −1×10−2 C - 1 \times {10^{ - 2}}\,{\text{C}} for q1{q_1} and 5×10−2 C5 \times {10^{ - 2}}\,{\text{C}} for q2{q_2} in the above equation.
q=(−1×10−2 C)+(5×10−2 C)q = \left( { - 1 \times {{10}^{ - 2}}\,{\text{C}}} \right) + \left( {5 \times {{10}^{ - 2}}\,{\text{C}}} \right)
⇒q=4×10−2 C\Rightarrow q = 4 \times {10^{ - 2}}\,{\text{C}}
Hence, the resultant charge on the two metallic spheres is 4×10−2 C4 \times {10^{ - 2}}\,{\text{C}}.
Substitute kq1r1\dfrac{{k{q_1}}}{{{r_1}}} for V1{V_1} and kq2r2\dfrac{{k{q_2}}}{{{r_2}}} for V2{V_2} in equation (2).
kq1r1=kq2r2\dfrac{{k{q_1}}}{{{r_1}}} = \dfrac{{k{q_2}}}{{{r_2}}}
⇒q1q2=r1r2\Rightarrow \dfrac{{{q_1}}}{{{q_2}}} = \dfrac{{{r_1}}}{{{r_2}}}
Substitute q−q2q - {q_2} for q1{q_1}, 1 cm1\,{\text{cm}} for r1{r_1} and 3 cm3\,{\text{cm}} for r2{r_2} in the above equation.
⇒q−q2q2=1 cm3 cm\Rightarrow \dfrac{{q - {q_2}}}{{{q_2}}} = \dfrac{{1\,{\text{cm}}}}{{3\,{\text{cm}}}}
⇒3(q−q2)=q2\Rightarrow 3\left( {q - {q_2}} \right) = {q_2}
⇒3q−3q2=q2\Rightarrow 3q - 3{q_2} = {q_2}
⇒q2+3q2=3q\Rightarrow {q_2} + 3{q_2} = 3q
⇒4q2=3q\Rightarrow 4{q_2} = 3q
⇒q2=34q\Rightarrow {q_2} = \dfrac{3}{4}q
Substitute 4×10−2 C4 \times {10^{ - 2}}\,{\text{C}} for qq in the above equation.
⇒q2=34(4×10−2 C)\Rightarrow {q_2} = \dfrac{3}{4}\left( {4 \times {{10}^{ - 2}}\,{\text{C}}} \right)
∴q2=3×10−2 C\therefore {q_2} = 3 \times {10^{ - 2}}\,{\text{C}}

Therefore, the charge on the bigger sphere is 3×10−2 C3 \times {10^{ - 2}}\,{\text{C}}.Hence, the correct option is A.

Note: One can also solve the same question by another method. One can determine the resultant potential and the resultant charge on both the metallic spheres and then using this resultant potential and the charge one can determine the charge on the bigger sphere which will be the required final answer.