Question

A proton is released in a uniform electric field, and it experiences an electric force of 2.04×10−14 N toward the south.

What is the magnitude of the electric field?
What is the direction of the electric field?
a. toward the north
b.toward the south
c. toward the east
d. toward the west

Answer 1

Part A:

`E=127500N/C\\E=1.275*10^(5) N/C`

Part B:

Option B (Towards the South)

Step-by-step explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

`E=(2.04*10^(-14))/(1.6*10^(-19)) \\E=127500N/C\\E=1.275*10^(5) N/C`

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.