Question

An object with a charge of -3.4 μC and a mass of 4.2×10−2 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. Find the magnitude of the electric field. If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.

Answer 1

(a)

Magnitude of the electric field can be calculated from the balancing condition, that is the weight of the object is balanced by the electrostatic force

If E is the electric field, the balancing equation is

mg=QE

Magnitude of electricfield

E=mg/Q

= 4.2∗10^−2∗9.8 ÷ (3.4∗10^−6)

E=121058.824 V/m

(b)

Charge on the object is doubled and mass remains the same

Force acting on the charged particle

F2=2*QE

=2∗3.4∗10^−6∗121058

=0.8232N

Magnitude of the acceleration a=F2/m

=0.8232/4.2∗10−2

a=19.6m/s2

Direction of acceleration will be in the direction of the force. For a negatively charged particle the force due the field will be in a direction opposite to the direction of the field. Electric field is directed downwards, so the force and acceleration is directed upwards.

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