Question

Let σa, σb, σc, and σd be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to be positive for now.
A: or A I E 0 B: on C: Oc D: OD
If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. Given that the electric field E⃗ due to a charged sheet with surface charge +σ is given by E=σ2ϵ0, and that it points away from the plane of the sheet, how can the condition that the electric field in plate I vanishes be written? Choose a option below.
a) σa−σb−σc−σd=0
b) −σa+σb−σc−σd=0
c) −σa−σb+σc−σd=0
d) −σa−σb−σc+σd=0

Answer 1

Answer:

a) σa−σb−σc−σd=0

Explanation:

The parallel plate capacitor is the one in which two metal plates are connected in parallel with some distancing among them. The electric field from both plates is denoted by E = σ / 2ϵ0. The σ is the charge density. The Electric field in plate I will vanish when the surface charge of σa is positive and rest of the charges are negative. The correct option is a.