Two large circular metal plates are parallel and nearly touching, only 4.1 mm apart. The two plates are connected to the opposite terminals of a 9 V battery.
(a) What is the average electric field strength in the space between the plates?
V/m
(b) What is the electric force on a 15 nC charge located halfway between the two plates? Give your answer in μN.
To answer these questions, we can use the formulas related to the electric field strength and the electric force between parallel plates.
(a) The formula for the electric field strength between two parallel plates is given by:
E = V/d,
where E is the electric field strength in volts per meter (V/m), V is the voltage between the plates in volts (V), and d is the distance between the plates in meters (m).
In this case, the voltage between the plates (V) is 9 V and the distance between the plates (d) is 4.1 mm, or 0.0041 m. Plugging these values into the formula, we get:
E = 9 V / 0.0041 m = 2195.12 V/m.
Therefore, the average electric field strength between the plates is approximately 2195.12 V/m.
(b) The formula for the electric force between two plates on a charge is given by:
F = qE,
where F is the electric force in newtons (N), q is the charge in coulombs (C), and E is the electric field strength in volts per meter (V/m).
In this case, the charge (q) is 15 nC, or 15 × 10^-9 C. The electric field strength (E) has already been calculated as 2195.12 V/m.
Plugging these values into the formula, we get:
F = (15 × 10^-9 C) × (2195.12 V/m) = 32.9278 × 10^-6 N.
Since 1 μN (micronewton) is equal to 10^-6 N, we can convert the answer to micro-newtons:
F = 32.9278 μN.
Therefore, the electric force on a 15 nC charge located halfway between the two plates is approximately 32.9278 μN.