Two sphere of radius 20 cm and 40 cm are charge to potential of 50v an...
Calculation of Common Potential and Loss of Energy
Given:
- Radius of first sphere = 20 cm
- Radius of second sphere = 40 cm
- Potential of first sphere = 50 V
- Potential of second sphere = 35 V
Calculation of Common Potential
When two spheres are connected by a wire, they attain a common potential. This potential can be calculated using the formula:
Vc = (Q1 + Q2) / (4πε0R)
- Q1 = 4πε0R12V1
- Q2 = 4πε0R22V2
Substituting the given values, we get:
- Q1 = 1.13 x 10-8 C
- Q2 = 1.58 x 10-7 C
- R1 = 20 cm = 0.2 m
- R2 = 40 cm = 0.4 m
- V1 = 50 V
- V2 = 35 V
- ε0 = 8.85 x 10-12 F/m
Using the above values in the formula, we get:
Vc = 43.13 V
Calculation of Loss of Energy
When the two spheres are connected by a wire, the charges will flow from the sphere with higher potential to the sphere with lower potential until they both attain the same potential. During this process, some energy will be lost in the form of heat due to the resistance of the wire. The loss of energy can be calculated using the formula:
E = (Q1 + Q2) / 2 x (V1 - V2)
Substituting the given values, we get:
- Q1 = 1.13 x 10-8 C
- Q2 = 1.58 x 10-7 C
- V1 = 50 V
- V2