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

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.

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explanation:

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it