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A uniform 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?

Respuesta :

Answer:

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

[tex]\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J[/tex]

ΔU =18.79 mJ