Two hollow conducting concentric spheres of radii R and r have positive charges with equal surface charge densities (σ). What is the electric potential at their common center? (r>R)

Question

Asked on June 1, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-01T02:01:17.264Z

Step 1: Identify the given parameters and the formula for electric potential. The inner conducting sphere has radius R_1 = 2 m and charge Q_1 = Q. The outer conducting sphere has radius R_2 = 6 m and charge Q_2 = -Q. The electric potential V at a distance r from the center of a conducting sphere with charge Q' and radius R' is given by: V(r) = (kQ')/(R') & for r R' \\ (kQ')/(r) & for r > R' where k = (1)/(4_0) is Coulomb's constant. Step 2: Calculate the electric potential on the surface of the inner sphere (V_1). The potential on the inner sphere (r=R_1) is due to its own charge Q_1 and the charge Q_2 on the outer sphere. The point r=R_1 is on the surface of the inner sphere and inside the outer sphere. Potential due to Q_1: V_1,Q_1 = (kQ_1)/(R_1) = (kQ)/(2) Potential due to Q_2: Since r=R_1 < R_2, the potential due to the outer sphere at this point is the same as its surface potential. V_1,Q_2 = (kQ_2)/(R_2) = (k(-Q))/(6) = -(kQ)/(6) The total potential on the inner sphere is: V_1 = V_1,Q_1 + V_1,Q_2 = (kQ)/(2) - (kQ)/(6) = kQ ( (1)/(2) - (1)/(6) ) = kQ ( (3-1)/(6) ) = kQ ( (2)/(6) ) = (kQ)/(3) Step 3: Calculate the electric potential on the surface of the outer sphere (V_2). The potential on the outer sphere (r=R_2) is due to the charge Q_1 on the inner sphere and its own charge Q_2. The point r=R_2 is outside the inner sphere and on the surface of the outer sphere. Potential due to Q_1: Since r=R_2 > R_1, V_2,Q_1 = (kQ_1)/(R_2) = (kQ)/(6) Potential due to Q_2: V_2,Q_2 = (kQ_2)/(R_2) = (k(-Q))/(6) = -(kQ)/(6) The total potential on the outer sphere is: V_2 = V_2,Q_1 + V_2,Q_2 = (kQ)/(6) - (kQ)/(6) = 0 Step 4: Calculate the potential difference between the two spheres. The potential difference between the spheres is V = V_1 - V_2. V = (kQ)/(3) - 0 = (kQ)/(3) Step 5: Compare the result with the given options. The calculated potential difference is (kQ)/(3), which matches option (2). The final answer is (2) (kQ)/(3). Send me the next one 📸

Accepted Answer · 2026-06-01T02:01:17.264Z

Step 1: Identify the given parameters and the formula for electric potential. The inner conducting sphere has radius R_1 = 2 m and charge Q_1 = Q. The outer conducting sphere has radius R_2 = 6 m and charge Q_2 = -Q. The electric potential V at a distance r from the center of a conducting sphere with charge Q' and radius R' is given by: V(r) = (kQ')/(R') & for r R' \\ (kQ')/(r) & for r > R' where k = (1)/(4_0) is Coulomb's constant. Step 2: Calculate the electric potential on the surface of the inner sphere (V_1). The potential on the inner sphere (r=R_1) is due to its own charge Q_1 and the charge Q_2 on the outer sphere. The point r=R_1 is on the surface of the inner sphere and inside the outer sphere. Potential due to Q_1: V_1,Q_1 = (kQ_1)/(R_1) = (kQ)/(2) Potential due to Q_2: Since r=R_1 < R_2, the potential due to the outer sphere at this point is the same as its surface potential. V_1,Q_2 = (kQ_2)/(R_2) = (k(-Q))/(6) = -(kQ)/(6) The total potential on the inner sphere is: V_1 = V_1,Q_1 + V_1,Q_2 = (kQ)/(2) - (kQ)/(6) = kQ ( (1)/(2) - (1)/(6) ) = kQ ( (3-1)/(6) ) = kQ ( (2)/(6) ) = (kQ)/(3) Step 3: Calculate the electric potential on the surface of the outer sphere (V_2). The potential on the outer sphere (r=R_2) is due to the charge Q_1 on the inner sphere and its own charge Q_2. The point r=R_2 is outside the inner sphere and on the surface of the outer sphere. Potential due to Q_1: Since r=R_2 > R_1, V_2,Q_1 = (kQ_1)/(R_2) = (kQ)/(6) Potential due to Q_2: V_2,Q_2 = (kQ_2)/(R_2) = (k(-Q))/(6) = -(kQ)/(6) The total potential on the outer sphere is: V_2 = V_2,Q_1 + V_2,Q_2 = (kQ)/(6) - (kQ)/(6) = 0 Step 4: Calculate the potential difference between the two spheres. The potential difference between the spheres is V = V_1 - V_2. V = (kQ)/(3) - 0 = (kQ)/(3) Step 5: Compare the result with the given options. The calculated potential difference is (kQ)/(3), which matches option (2). The final answer is (2) (kQ)/(3). Send me the next one 📸

Two hollow conducting concentric spheres of radii R and r have positive charges with equal surface charge densities (σ). What is the electric potential at their common center? (r>R)

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Two hollow conducting concentric spheres of radii R and r have positive charges with equal surface charge densities (σ). What is the electric potential at their common center? (r>R)

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