The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r² Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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