Calculate the electric field strength if the potential across charged plates is 12 volts and their separation is 0.4 cm. 5v/m 0.05 v/m 30 v/m 3,000 v/m calculate the force a test charge of 0.001 coulomb would experience in the field across charged plates of 12 volts and separation of 0.4 cm. 5 times 10-^3 n 5 times 10-^5 n 5 times 10 -^2n 3

Answer :

(1) The electric field strength is, 3,000 V/m

(2) The force a test charge is, 3 N

Solution for part 1 :

Formula used for electric field strength :

where,

E = electric field strength

V = potential = 12 volts

d = distance between the charge = 0.4 cm = 0.004 m     (1 m = 100 cm)

Now put all the given values in the above formula, we get the electric field strength.

Therefore, the electric field strength is, 3000 V/m

Solution for part 2 :

Another formula used for electric field strength with force :

where,

E = electric field strength = 3000 V/m     (from the solution of part 1)

F = force

q = charge = 0.001 coulomb

Now put all the given values in the above formula, we get the force a test charge.

Therefore, the force a test charge is, 3 N

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

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