Two charged metallic spheres with radii \(R_1\) and \(R_2\) are brought in contact and then separated. The ratio of final charges \(Q_1\) and \(Q_2\) on the two spheres respectively will be _____.
Fill in the blank with the correct answer from the options given below
Fill in the blank with the correct answer from the options given below
- \(\frac{Q_2}{Q_1}=\frac{R_1}{R_2}\)
- \(\frac{Q_2}{Q_1}<\frac{R_1}{R_2}\)
- \(\frac{Q_2}{Q_1}>\frac{R_1}{R_2}\)
- \(\frac{Q_2}{Q_1}=\frac{R_2}{R_1}\)
The Correct Option is D
Solution and Explanation
The correct answer is Option 4: Q1/Q2 = R1/R2 or Q2/Q1 = R2/R1.
Here's the explanation:
When two charged metallic spheres are brought into contact, they reach an equilibrium state where their potentials are equal. This is because charge flows from the sphere with higher potential to the sphere with lower potential until the potentials are the same.
The potential (V) of a charged sphere is given by:
V = kQ/R
Where:
- V is the potential
- k is Coulomb's constant
- Q is the charge on the sphere
- R is the radius of the sphere
When the spheres are in contact, their potentials are equal:
V1 = V2
Therefore:
kQ1/R1 = kQ2/R2
We can cancel k from both sides:
Q1/R1 = Q2/R2
Rearranging to find the ratio of charges:
Q1/Q2 = R1/R2
Q2/Q1 = R2/R1
Thus, the ratio of the final charges on the two spheres is directly proportional to the ratio of their radii.
Therefore, the correct answer is:
Option 4: Q2/Q1 = R2/R1
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