Two metallic solid spheres A and B,have radius R and 3R,respectively. The solid spheres are charged and kept isolated. Then,the two spheres are connected to each other through a thin conducting wire. The ratio of the final charge on the spheres A to B is:
1:1
1:3
3:1
1:9
9:1
The Correct Option is B
Approach Solution - 1
Two isolated charged spheres with radii R and 3R are connected by a conducting wire.
When connected, the potentials equalize: \[ V_A = V_B \] \[ \frac{kQ_A}{R} = \frac{kQ_B}{3R} \]
Charge Ratio is \[ \frac{Q_A}{Q_B} = \frac{R}{3R} = \frac{1}{3} \]
Alternatively, since capacitance C ∝ radius: \[ \frac{C_A}{C_B} = \frac{R}{3R} = \frac{1}{3} \] And Q = CV, so when V equalizes: \[ \frac{Q_A}{Q_B} = \frac{C_A}{C_B} = \frac{1}{3} \]
Approach Solution -2
1. Understand the concept of charge distribution:
When two charged conductors are connected by a conducting wire, charge flows between them until they reach the same potential.
2. Recall the formula for the potential of a sphere:
The electric potential (V) of a conducting sphere with charge Q and radius R is given by:
\[V = \frac{kQ}{R}\]
where k is Coulomb's constant.
3. Set the final potentials equal:
Let \(Q_A\) and \(Q_B\) be the initial charges on spheres A and B, and \(Q_A'\) and \(Q_B'\) be their final charges after they are connected. Since the spheres reach the same potential after connection:
\[V_A' = V_B'\]
\[\frac{kQ_A'}{R} = \frac{kQ_B'}{3R}\]
4. Solve for the charge ratio:
Simplify the equation:
\[Q_A' = \frac{Q_B'}{3}\]
\[\frac{Q_A'}{Q_B'} = \frac{1}{3}\]
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).