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A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surfaces of solid sphere and hollow shell is V. If the shell is now given a charge of -3Q, then the new potential difference between the two surfaces will be
- a)V
- b)2 V
- c)4 V
- d)-2 V
Correct answer is option 'A'. Can you explain this answer?
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A solid conducting sphere having a charge Q is surrounded by an unchar...
Potential difference between surfaces of solid sphere and hollow shell
When the solid conducting sphere is charged, it creates an electric field around it. The potential difference between the surface of the solid sphere and the hollow shell is determined by the electric field created by the charge on the solid sphere.
When the solid sphere is uncharged, the electric field inside it is zero. This means that the potential inside the solid sphere is constant, and equal to the potential at its surface. Let's call this potential V_s.
The potential outside the solid sphere is determined by the charge Q on the sphere and the distance r from its center. By using the formula for electric potential of a point charge, we can write:
V = k * Q / r
where k is the electrostatic constant.
Since the hollow shell is uncharged, there is no electric field inside it. Therefore, the potential inside the hollow shell is also constant, and equal to the potential at its surface. Let's call this potential V_h.
When the hollow shell is given a charge of -3Q, it creates an electric field around it. This electric field will affect the potential inside the hollow shell.
New potential difference between the two surfaces
Since the potential inside the solid sphere is constant, it remains V_s.
The potential outside the hollow shell is determined by the charge on the solid sphere and the charge on the hollow shell. By using the principle of superposition, we can write:
V' = k * (Q - 3Q) / r
Simplifying the expression:
V' = -2k * Q / r
Since the potential inside the hollow shell is constant, it remains V_h.
Therefore, the new potential difference between the two surfaces is:
V' - V_h = -2k * Q / r - V_h
Since V_h = V, we can substitute it into the expression:
V' - V = -2k * Q / r - V
V' - V = -2V
Therefore, the new potential difference between the two surfaces is -2V, which is option D.
When the solid conducting sphere is charged, it creates an electric field around it. The potential difference between the surface of the solid sphere and the hollow shell is determined by the electric field created by the charge on the solid sphere.
When the solid sphere is uncharged, the electric field inside it is zero. This means that the potential inside the solid sphere is constant, and equal to the potential at its surface. Let's call this potential V_s.
The potential outside the solid sphere is determined by the charge Q on the sphere and the distance r from its center. By using the formula for electric potential of a point charge, we can write:
V = k * Q / r
where k is the electrostatic constant.
Since the hollow shell is uncharged, there is no electric field inside it. Therefore, the potential inside the hollow shell is also constant, and equal to the potential at its surface. Let's call this potential V_h.
When the hollow shell is given a charge of -3Q, it creates an electric field around it. This electric field will affect the potential inside the hollow shell.
New potential difference between the two surfaces
Since the potential inside the solid sphere is constant, it remains V_s.
The potential outside the hollow shell is determined by the charge on the solid sphere and the charge on the hollow shell. By using the principle of superposition, we can write:
V' = k * (Q - 3Q) / r
Simplifying the expression:
V' = -2k * Q / r
Since the potential inside the hollow shell is constant, it remains V_h.
Therefore, the new potential difference between the two surfaces is:
V' - V_h = -2k * Q / r - V_h
Since V_h = V, we can substitute it into the expression:
V' - V = -2k * Q / r - V
V' - V = -2V
Therefore, the new potential difference between the two surfaces is -2V, which is option D.
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A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surfaces of solid sphere and hollow shell is V. If the shell is now given a charge of -3Q, then the new potential difference between the two surfaces will bea)Vb)2 Vc)4 Vd)-2 VCorrect answer is option 'A'. Can you explain this answer?
Question Description
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surfaces of solid sphere and hollow shell is V. If the shell is now given a charge of -3Q, then the new potential difference between the two surfaces will bea)Vb)2 Vc)4 Vd)-2 VCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surfaces of solid sphere and hollow shell is V. If the shell is now given a charge of -3Q, then the new potential difference between the two surfaces will bea)Vb)2 Vc)4 Vd)-2 VCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surfaces of solid sphere and hollow shell is V. If the shell is now given a charge of -3Q, then the new potential difference between the two surfaces will bea)Vb)2 Vc)4 Vd)-2 VCorrect answer is option 'A'. Can you explain this answer?.
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