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If all the excess charge Q is transferred from a conducting sphere of radius R at a surface potential of 30V to an originally neutral sphere of twice the radius (2R), what is the new sphere's surface potential? O 15/2 V O 15V O 60V O 30V O 18V

If all the excess charge Q is transferred from a conducting sphere of radius R at a surface potential of 30V to an originally neutral sphere of twice the radius (2R), what is the new sphere's surface potential? O 15/2 V O 15V O 60V O 30V O 18V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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### Physics Problem: Surface Potential of Spheres **Problem Statement:** If all the excess charge \( Q \) is transferred from a conducting sphere of radius \( R \) at a surface potential of 30V to an originally neutral sphere of twice the radius \( (2R) \), what is the new sphere's surface potential? **Options:** - \( \dfrac{15}{2} \)V - 15V - 60V - 30V - 18V **Explanation:** This question involves understanding the relationship between the surface potential of conducting spheres, their radii, and the charge they carry. The surface potential \( V \) of a conducting sphere with charge \( Q \) and radius \( R \) can be described using the formula: \[ V = \dfrac{Q}{4 \pi \varepsilon_0 R} \] Here's a step-by-step approach to solving the problem: 1. **Given:** - Initial sphere radius \( R \). - Initial surface potential \( V_1 = 30V \). - Charge \( Q \) on the sphere. - New sphere radius \( 2R \). 2. **Calculate the initial charge \( Q \):** - Using the initial sphere formula: \[ 30V = \dfrac{Q}{4 \pi \varepsilon_0 R} \] - Rearrange to solve for \( Q \): \[ Q = 30V \times 4 \pi \varepsilon_0 R \] 3. **Transfer the charge \( Q \) to a new sphere with radius \( 2R \):** - The new surface potential \( V_2 \) can be calculated using the same formula for the new sphere: \[ V_2 = \dfrac{Q}{4 \pi \varepsilon_0 (2R)} \] - Substitute \( Q \) from the initial sphere: \[ V_2 = \dfrac{30V \times 4 \pi \varepsilon_0 R}{4 \pi \varepsilon_0 \times 2R} \] - Simplify the expression: \[ V_2 = \dfrac{30V}{2} \] \[ V_2 = 15V \] **Therefore, the new sphere's surface
Transcribed Image Text:### Physics Problem: Surface Potential of Spheres **Problem Statement:** If all the excess charge \( Q \) is transferred from a conducting sphere of radius \( R \) at a surface potential of 30V to an originally neutral sphere of twice the radius \( (2R) \), what is the new sphere's surface potential? **Options:** - \( \dfrac{15}{2} \)V - 15V - 60V - 30V - 18V **Explanation:** This question involves understanding the relationship between the surface potential of conducting spheres, their radii, and the charge they carry. The surface potential \( V \) of a conducting sphere with charge \( Q \) and radius \( R \) can be described using the formula: \[ V = \dfrac{Q}{4 \pi \varepsilon_0 R} \] Here's a step-by-step approach to solving the problem: 1. **Given:** - Initial sphere radius \( R \). - Initial surface potential \( V_1 = 30V \). - Charge \( Q \) on the sphere. - New sphere radius \( 2R \). 2. **Calculate the initial charge \( Q \):** - Using the initial sphere formula: \[ 30V = \dfrac{Q}{4 \pi \varepsilon_0 R} \] - Rearrange to solve for \( Q \): \[ Q = 30V \times 4 \pi \varepsilon_0 R \] 3. **Transfer the charge \( Q \) to a new sphere with radius \( 2R \):** - The new surface potential \( V_2 \) can be calculated using the same formula for the new sphere: \[ V_2 = \dfrac{Q}{4 \pi \varepsilon_0 (2R)} \] - Substitute \( Q \) from the initial sphere: \[ V_2 = \dfrac{30V \times 4 \pi \varepsilon_0 R}{4 \pi \varepsilon_0 \times 2R} \] - Simplify the expression: \[ V_2 = \dfrac{30V}{2} \] \[ V_2 = 15V \] **Therefore, the new sphere's surface
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