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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

Respuesta :

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 :

[tex]E=\frac{V}{d}[/tex]

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.

[tex]E=\frac{12volts}{0.004m}=3000V/m[/tex]

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

Solution for part 2 :

Another formula used for electric field strength with force :

[tex]E=\frac{F}{q}[/tex]

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.

[tex]3000V/m=\frac{F}{0.001coulomb}[/tex]

[tex]F=3N[/tex]

Therefore, the force a test charge is, 3 N

Answer:

3,000 V/m

3

Explanation:

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