It is the amount of work done to bring a charge q from infinity to that point against the electric fields of a given charge without changing its kinetic energy.
PE = qQ/4πε0r = qV
Work done W = ?PE = PE final - PE initial
= q1q2/4πε0 [(1/rf) - (1/r1]
Force on a surface charge:- the repulsive force acting on an element of a charged surface due to rest of the charged surface is called electric force on a charged conducting surface.
E = E! + E2 = σ/ε0 where E1 is the electric field due to small element and E2 is due to rest of the charge.
E1 = E2
Therefore electric field intensity near a charged surface = σ/ 2ε0 and force dF = σ2 / 2ε0 ds
Therefore F = ?σ2 / 2ε0 ds
And pressure p elect. = dF / ds = σ2 / 2ε0 = ε0 E2 / 2
In case of a soap bubble
P in - p out = P excess = P ST = P elect. = 4T / r = σ2 / 2ε0
= 4T / r - q2 / 2 A2 ε20 = 4T / r - q2 / 32 π2 r4 ε0
If air pressure inside and outside the bubble are equal then
Pin = Pout or 4T / r = q2 / 32 π2 r4 ε0
Energy density = energy per unit volume
U = ε0 E 2 / 2 = σ2 / 2ε0
Total energy U tot = ∫ε0 E2 / 2 dV
Charged liquid drop if a identical drops each of radius r and charge q are joined to form a big drop of radius R and charge Q
Then R = n1/3 r Q big = nq small
E big = n1/3 E small; Vbig = n2/3 V small; σ big = n1/3 σ small
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