
Two metallic spheres of radii 1cm and 3cm are given charges of $ - 1 \times {10^{ - 2}}C$ and $5 \times {10^{ - 2}}C$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
A) $3 \times {10^{ - 2}}C$
B) $4 \times {10^{ - 2}}C$
C) $1 \times {10^{ - 2}}C$
D) $2 \times {10^{ - 2}}C$
Answer
571.8k+ views
Hint: When two charged bodies are connected with each other with some conducting rod, the flow of charge takes place between the two bodies until the potential of both the bodies becomes the same. In these types of questions, equate the final potential of both bodies. Also, use the fact that the net charge of the system is conserved.
Complete step by step solution:
When two charged spheres are connected with each other with conducting wire, the flow of charge will take place between the two spheres until the potential of both spheres becomes the same.
Let there be $x$ charge on the bigger sphere after the flow of charge finally stops.
But the total charge of the two system will remain conserves as $\left( { - 1 \times {{10}^{ - 2}}} \right) + \left( {5 \times {{10}^{ - 2}}} \right) = 4 \times {10^{ - 2}}C$
The charge on smaller sphere will be $(4 \times {10^{ - 2}} - x)C$
The potential of a spherically charged distribution is given by the formula $V = k\dfrac{q}{r}$
Let the final potential of bigger sphere be ${V_1}$ and that of smaller sphere be ${V_2}$
${V_1} = k\dfrac{x}{3}$
${V_2} = k\dfrac{{\left( {4 \times {{10}^{ - 2}} - x} \right)}}{1}$
Equating both the potentials of bigger and smaller blocks,
${V_1} = {V_2}$
$k\dfrac{x}{3} = k\dfrac{{(4 \times {{10}^{ - 2}} - x)}}{1}$
$x = 12 \times {10^{ - 2}} - 3x$
$x = 3 \times {10^{ - 2}}C$
$\therefore $ The final charge on the bigger sphere is $3 \times {10^{ - 2}}C$. So, option (A) is correct.
Note:
If the charged sphere is connected to the earth’s surface then it will be known as earthing.
The process of transferring the immediate discharge of the electrical energy directly to the earth by the help of the low resistance wire is known as the electrical earthing. The electrical earthing is done by connecting the non-current carrying part of the equipment or neutral of the supply system to the ground.
Complete step by step solution:
When two charged spheres are connected with each other with conducting wire, the flow of charge will take place between the two spheres until the potential of both spheres becomes the same.
Let there be $x$ charge on the bigger sphere after the flow of charge finally stops.
But the total charge of the two system will remain conserves as $\left( { - 1 \times {{10}^{ - 2}}} \right) + \left( {5 \times {{10}^{ - 2}}} \right) = 4 \times {10^{ - 2}}C$
The charge on smaller sphere will be $(4 \times {10^{ - 2}} - x)C$
The potential of a spherically charged distribution is given by the formula $V = k\dfrac{q}{r}$
Let the final potential of bigger sphere be ${V_1}$ and that of smaller sphere be ${V_2}$
${V_1} = k\dfrac{x}{3}$
${V_2} = k\dfrac{{\left( {4 \times {{10}^{ - 2}} - x} \right)}}{1}$
Equating both the potentials of bigger and smaller blocks,
${V_1} = {V_2}$
$k\dfrac{x}{3} = k\dfrac{{(4 \times {{10}^{ - 2}} - x)}}{1}$
$x = 12 \times {10^{ - 2}} - 3x$
$x = 3 \times {10^{ - 2}}C$
$\therefore $ The final charge on the bigger sphere is $3 \times {10^{ - 2}}C$. So, option (A) is correct.
Note:
If the charged sphere is connected to the earth’s surface then it will be known as earthing.
The process of transferring the immediate discharge of the electrical energy directly to the earth by the help of the low resistance wire is known as the electrical earthing. The electrical earthing is done by connecting the non-current carrying part of the equipment or neutral of the supply system to the ground.
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