During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1), lightning will occur. If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground. In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?

Electric field between the cloud and ground,V = E.dV= 3 x 106 x 1000 = 3 x 109V Electrons transfered from cloud to ground,q = 25 CElectron static potential energy,U = `1/2 "CV"^2``["C" = "q"/"V"]``= 1/2 "qV" = 1/2 xx 25 xx 3 xx 10^9` U = 37.5 x 109 J

During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. - Physics

Question

During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 10⁶ Vm^–1), lightning will occur.

If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?

Long Answer

Solution

Electric field between the cloud and ground,
V = E.d
V= 3 x 10⁶ x 1000 = 3 x 10⁹V
Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = `1/2 "CV"^2`
`["C" = "q"/"V"]`
`= 1/2 "qV" = 1/2 xx 25 xx 3 xx 10^9`

U = 37.5 x 10⁹ J

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Q 13.Q 12.Q 14.

Chapter 1: Electrostatics - Evaluation [Page 78]

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Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 12 TN Board

Chapter 1 Electrostatics
Evaluation | Q 13. | Page 78

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