Q15. A cube of side 'a' has a charge Q at each of its vertices, what is the potential due to this ch

Q: What is the correct answer for CUET UG Physics Q15?

The correct answer is b. To find the potential at the center of the cube, we first need to determine the distance from the center to any vertex. A cube has 8 vertices, and since the electric potential is a scalar quantity, we can sum the potentials due to each charge.Let the side length of the cube be 'a'.The length of the main diagonal of the cube (connecting opposite vertices) is L = √(a² + a² + a²) = a√3.The distance f

Q: Q15. A cube of side 'a' has a charge Q at each of its vertices, what is the potential due to this charge array at the centre of the cube? (a) 4Q / (3πε₀a) (b) Q / (√3πε₀a) (c) 2Q / (√3πε₀a) (d) 8Q /…

The correct answer is b. To find the potential at the center of the cube, we first need to determine the distance from the center to any vertex. A cube has 8 vertices, and since the electric potential is a scalar quantity, we can sum the potentials due to each charge.Let the side length of the cube be 'a'.The length of the main diagonal of the cube (connecting opposite vertices) is L = √(a² + a² + a²) = a√3.The distance f

Q: What are the options for this CUET UG Physics question?

A. 4Q / (3πε₀a) · B. Q / (√3πε₀a) · C. 2Q / (√3πε₀a) · D. 8Q / (πε₀a)

Question

Q15. A cube of side 'a' has a charge Q at each of its vertices, what is the potential due to this charge array at the centre of the cube?
(a) 4Q / (3πε₀a)
(b) Q / (√3πε₀a)
(c) 2Q / (√3πε₀a)
(d) 8Q / (πε₀a)

FirstInTest · Accepted Answer

To find the potential at the center of the cube, we first need to determine the distance from the center to any vertex. A cube has 8 vertices, and since the electric potential is a scalar quantity, we can sum the potentials due to each charge.Let the side length of the cube be 'a'.The length of the main diagonal of the cube (connecting opposite vertices) is L = √(a² + a² + a²) = a√3.The distance from the center of the cube to any vertex (r) is half of the main diagonal: r = L/2 = (a√3)/2.The potential (V) at a distance 'r' from a single point charge 'Q' is given by V_single = Q / (4πε₀r).Since there are 8 charges, each of magnitude Q, and all are equidistant from the center, the total potential (V_total) at the center is:V_total = 8 × V_single = 8 × [Q / (4πε₀r)]V_total = 8Q / (4πε₀ × (a√3)/2)V_total = 8Q × 2 / (4πε₀a√3)V_total = 16Q / (4πε₀a√3)V_total = 4Q / (πε₀a√3).However, this calculated value (4Q / (πε₀a√3)) does not directly match any of the given options exactly. The provided correct answer is (b) Q / (√3πε₀a). This value would correspond to the potential if the distance from the center to a vertex was (a√3)/4, or if there was only one charge Q at a distance (a√3)/2 and the formula was V = Q / (2πε₀r). There seems to be a discrepancy between the standard physics calculation and the provided answer choice/values. If option (b) were to be correct based on the typical formula, it would mean that the total potential is 1/4th of the correctly calculated value, or that Q is actually 1/4th of the charge required for the calculation.

Answer

4Q / (3πε₀a)

Answer

Q / (√3πε₀a)

Answer

2Q / (√3πε₀a)

Answer

8Q / (πε₀a)

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