On the diagram of the charge distributtion below, sketch the electric field in t

Question

On the diagram of the charge distributtion below, sketch the electric field in the region all around the charges. a) Draw about 20 lines or so, giving a good overview of their density and direction. b) Draw in few equipotential surfaces. The two plates parallel plate capacitor have a potential difference of 25 V between them, and are separated by a distance of 10 mum. What is the magnitude of the electric field between them? What would be the change in electric potential energy if a charge of +3e is accelerated between the plates? (Include sign.)

Explanation / Answer

here,

Conversion:
1m = 10^6 um

Potential Difference, V = 25 V
seperation distance, d = 10*10^-6 m

Part A:
Electric field, E = - v/d ----------(1)
E = -25 / (10*10^-6)
E = -2500000 V/m

Part B:
Change in Potential Energy, del(u) = q*(v)
del(u) = q*E*d (Using eqn 1 for V)
del(u) = 3 * -2500000 * (10*10^-6)
del(u) = -75 J

-negative sign is merealy just a direction

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