Question

What must the charge (sign and magnitude) of a 1.43-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 690 N/C ?

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Answer 1

a)

Simce the lectric field is acting downwards we would want a charge such that this electric field applies a force on it in the upward direction opposite to gravity. Hence the charge must be negative.

Now all we have to do us balance the gravity force mg to the Electric field force qE.

mass of the particle m =1.43 g =1.43*10^-3 kg

the electric field E =690(-j)N/C

(since electric field in downward direction)

magnitude of the electric field IEI =690 N/C

acceleration due to gravity g =9.81 m/s^2

the relation between electric field and electro static force is

                     F =qE

          charge q =F/E    ...... (1)

but , from Newton's second law of motion,

              force F =mg   ....... (2)

therefore, magnitude of charge q =mg/E   ....... (3)

                                                q =20.31*10^-6 C

                                                q =20.31 μC (negative)

.................................................

b)

from Newton's second law of motion,

              force F =mg   ....... (4)

the relation between electric field and electro static force is

                     F =qE ......... (5)

compare eq (4) and (5), we get

                  qE =mg

electric field E =mg/q   ....... (6)

where, mass of proton m =1.67*10^-27 kg

           charge of proton q =1.6*10^-19 C

           and g =9.81 m/s^2

magnitude of the electric field E =mg/q

                                                =10.23*10^-8 N/C