Question

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What are the magnitude and direction of the electric field?

Answer 1

The magnitude of the electric field is 6.53 × 10^5 N/C and the direction is North.

Step-by-step explanation:

The magnitude of the electric field can be determined using the equation of motion for accelerated motion. The equation is given as:

a = qE/m

Where:

• a is the acceleration
• q is the charge of the particle
• E is the electric field
• m is the mass of the particle

In this case, the acceleration is known to be 115 m/s² and the charge of an electron is -1.6 × 10^-19 C. The mass of an electron is approximately 9.11 × 10^-31 kg. Plugging in these values, we get:

a = (-1.6 × 10^-19 C)(E)/(9.11 × 10^-31 kg)

Simplifying the equation, we can solve for E:

E = (m * a) / q

Substituting the known values, we have:

E = (9.11 × 10^-31 kg)(115 m/s²) / (-1.6 × 10^-19 C) = -6.53 × 10^5 N/C

The magnitude of the electric field is 6.53 × 10^5 N/C and the direction is North, as given in the question.

Answer 2

The direction of the electric field would be south.

qE/m = 115
E = 115*m/q
= 115 * 9.1 * 10^(-31) / 1.67*10^(-19)
= 762.87 * 10^(-12)
= 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.