The potential on charging the surface of a conducting sphere of radius 10 cm is 15 V. What is the potential at its centre?
Answers
Answer:
When the two conducting spheres touch each other there will be a flow of charge until they both have the same potential.
Let R1 and R2 be the radii of spheres 1 and 2, respectively. Let Q1 and Q2 be the charges on spheres 1 and 2, respectively, after they are separated. Since both have the same potential,
V = Q1/C1 = Q2/C2
⇒ Q1/(4πϵ0R1) = Q2/(4πϵ0R2)
⇒ Q1/R1 = Q2/R2
⇒ Q1/Q2 = R1/R2
Surface charge density on sphere 1 = σ1 = Q1/(4πR12)
Surface charge density on sphere 2 = σ2 = Q2/(4πR22)
σ1 /σ2 = (Q1/Q2) x (R22/R12)
= (R1/R2) x (R22/R12)
= R2/R1
= 20 cm/10 cm
= 2
Therefore σ1 : σ2 = 2 : 1
Answer:
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Explanation:
Answer:
When the two conducting spheres touch each other there will be a flow of charge until they both have the same potential.
Let R1 and R2 be the radii of spheres 1 and 2, respectively. Let Q1 and Q2 be the charges on spheres 1 and 2, respectively, after they are separated. Since both have the same potential,
V = Q1/C1 = Q2/C2
⇒ Q1/(4πϵ0R1) = Q2/(4πϵ0R2)
⇒ Q1/R1 = Q2/R2
⇒ Q1/Q2 = R1/R2
Surface charge density on sphere 1 = σ1 = Q1/(4πR12)
Surface charge density on sphere 2 = σ2 = Q2/(4πR22)
σ1 /σ2 = (Q1/Q2) x (R22/R12)
= (R1/R2) x (R22/R12)
= R2/R1
= 20 cm/10 cm
= 2
Therefore σ1 : σ2 = 2 : 1