The E field strength between two charged parallel plates is the ratio of the potential difference and separation of the plates
Two parallel metal plates are separated by 3.5 cm and have a potential difference of 7.9 kV. Calculate the electric force acting on a stationary charged particle between the plates that has a charge of 2.6 × 10-15?C.
Step 1:?Write down the known values
Potential difference, ΔV = 7.9 kV = 7.9 × 103?V
Distance between plates, Δd = 3.5 cm = 3.5 × 10-2?m
Charge, Q = 2.6 × 10-15?C
Step 2:?Calculate the electric field strength between the parallel plates
Step 3:?Write out the equation for electric force on a charged particle
F = QE
Step 4:?Substitute electric field strength and charge into electric force equation
F = QE = (2.6 × 10-15) × (2.257 × 105) = 5.87 × 10-10?N = 5.9 × 10-10?N (2 s.f.)
A metal sphere of diameter 15 cm is negatively charged. The electric field strength at the surface of the sphere is 1.5 × 105?V m-1. Determine the total surface charge of the sphere.
Step 1:?Write down the known values
Electric field strength, E = 1.5 × 105?V m-1
Radius of sphere, r = 15 / 2 = 7.5 cm = 7.5 × 10-2?m
Step 2:?Write out the equation for electric field strength
Step 3:?Rearrange for charge Q
Q = 4πε0Er2
Step 4:?Substitute in values
Q = (4π × 8.85 × 10-12) × (1.5 × 105) × (7.5 × 10-2)2?= 9.38 × 10-8?C = 94 nC (2 s.f)
Remember to always?square?the distance!
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