Question

A charge q = 5 x 10-6 C starts at rest in a uniform electric field of 100 N/C. The electric field is parallel to the +x-direction.   How far has the particle traveled after 10 s? The mass of the charge is 5 mg.

Answer 1

Comment below if you have any query.

Relation between force and electric field is given by:

F = q*E

Since q = 5*10^-6 C (Charge is positive)

Since Electric field is towards right and charge in field is +ve, So force on charge will be in the same direction of electric field, which is towards +x-axis.

Now Using Force balance, From newton's 2nd law:

Fnet = m*a

So,

m*a = q*E

a = q*E/m

Now Using given values, acceleration will be

m = mass of charge = 5 mg = 5*10^-6 kg

E = 100 N/C

a = 5*10^-6*100/(5*10^-6)

a = 100 m/sec^2

Vi = Initial velocity = 0 m/sec

d = distance traveled by particle in 10 sec

Using 2nd kinematic equation

d = Vi*t + 0.5*a*t^2

d = 0*10 + 0.5*100*10^2

d = 5000 m = distance traveled by particle

d = 5*10^3 m = 5 km

Use the correct unit, which is required in your question (m or km). Let me know if you need anything else

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