Auniform electric field has a magnitude 1.80 kv/m and points in the +x direction. (a) what is the electric potential difference between x = 0.00 m plane and the x = 3.60 m plane? v(3.60 m) = kv (b) a point particle that has a charge of +2.90 µc is released from rest at the origin. what is the change in the electric potential energy of the particle as it travels from the x = 0.00 m plane to the x = 3.60 m plane?

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

ΔU =18.79 mJ

area of focus

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