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Assertion: In case of charged spherical ...

Assertion: In case of charged spherical shells, E-r graph is discontinuous while V-r graph is continuous
Reason: According to Gauss's theorem only the charge inside a closed surface ca produce electric field at some point.

A

If both assertion and reason are ture and reason is the correct explanation of assertion.

B

If both assertin and reason are ture but reason is not the correct explanation of assertion .

C

If assertion is true but reason is false.

D

If both assertion and reason are false.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze the assertion and reason provided and check their validity step by step. ### Step 1: Understanding the Assertion The assertion states that "In case of charged spherical shells, the E-r graph is discontinuous while the V-r graph is continuous." - **Electric Field (E)**: For a charged spherical shell, according to Gauss's law, the electric field inside the shell (at any point where r < R, where R is the radius of the shell) is zero. Outside the shell (at r > R), the electric field behaves as if all the charge were concentrated at the center. Therefore, the E-r graph shows a discontinuity at the surface of the shell. - **Electric Potential (V)**: The electric potential inside a charged spherical shell is constant (equal to the potential at the surface) and does not change until you reach the surface. After that, it decreases as you move away from the shell. Thus, the V-r graph is continuous. ### Step 2: Understanding the Reason The reason states: "According to Gauss's theorem, only the charge inside a closed surface can produce an electric field at some point." - This statement is partially correct. Gauss's law states that the electric field produced by a charge is determined by the charge enclosed within a Gaussian surface. However, it does not imply that charges outside the surface do not produce an electric field; they do, but their contributions are not included in the calculation of the electric field at points inside the Gaussian surface. ### Step 3: Conclusion - The assertion is **true** because the E-r graph is indeed discontinuous while the V-r graph is continuous for a charged spherical shell. - The reason is **false** because it incorrectly implies that charges outside the Gaussian surface do not contribute to the electric field at all. ### Final Answer Since the assertion is true and the reason is false, the correct answer is that the assertion is correct, but the reason is not a valid explanation for the assertion.
To solve the question, we need to analyze the assertion and reason provided and check their validity step by step. ### Step 1: Understanding the Assertion The assertion states that "In case of charged spherical shells, the E-r graph is discontinuous while the V-r graph is continuous." - **Electric Field (E)**: For a charged spherical shell, according to Gauss's law, the electric field inside the shell (at any point where r < R, where R is the radius of the shell) is zero. Outside the shell (at r > R), the electric field behaves as if all the charge were concentrated at the center. Therefore, the E-r graph shows a discontinuity at the surface of the shell. - **Electric Potential (V)**: The electric potential inside a charged spherical shell is constant (equal to the potential at the surface) and does not change until you reach the surface. After that, it decreases as you move away from the shell. Thus, the V-r graph is continuous. ...
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