Please please help me!!!!!! Also try to tell me that "another uncharged 600pF capacitor" is connected in series or parallel?
Answers
Answer:
6 x 10-6 J
Explanation:
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E = (1/2) CV²
= (1/2) * 600 × 10⁻¹² × (200)²
= 1.2 * 10⁻⁵ J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,
1/C'= (1/C) + (1/C)
= 1/600 + 1/600
= 1/300
C' = 300 pF
New electrostatic energy can be calculated as
E'= (1/2) * C' * V²
= (1/2) * 300 * 200²
= 0.6 * 10⁻⁵ J
Loss in electrostatic enegy = E - E'
= 1.2 x 10-5 - 0.6 x 10-5
= 0.6 x 10-5
= 6 x 10-6 J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
Hope it helps!
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E = 1/2CV²
= 1/2×600×10-¹²×(200)²
= 1.2×10^-5J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,
1/C' = (1/C)+(1/C)
= (1/600)+(1/600)
= (2/600)
= (1/300)
C' = 300pF
New electrostatic energy can be calculated as
E' = 1/2×C'×V²
= 1/2×300×200²
= 0.6×10^-5 J
Loss in electrostatic enegy = E - E'
= 1.2 x 10-5 - 0.6 x 10-5
= 0.6 x 10-5
= 6 x 10-6 J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
Hope it helps!!
Thank you ✌️