Question

What is the magnitude of an electric field in which the electric force it exerts on a proton is equal in magnitude to the proton's weight?

Answer 1

Answer:

5.58 × 10⁻¹¹ N/C

Explanation:

Electric force on a charge, F = qE

where,

F is the electric force in Newton (N)

q is the unit charge in Coulomb (C)

E is the electric field intensity in N/C

Also, the weight of an object, W = mg

Where,

w is weight of an object in N

m is mass in kg

g is acceleration due to gravity in m/s²

Now from the question given, the electric force exerted on the proton is equal in magnitude to the proton's weight.

                           ⇒     F = w

                                   qE = mg

making electric field, E the subject of the equation,

                                    $E = \frac{mg}{q}$

                             E = (9.11 × 10⁻³¹ kg × 9.8 m/s) ÷ 1.6 × 10⁻¹⁹ C

                             E =  55.8 × 10⁻¹² N/C

                             E =  5.58 × 10⁻¹¹ N/C