Physics: A charge of 7.2x10^-5C is placed in an Electric field with a strength of 4.8x10^5N/C. if the Electric potential energy of the charge is 75 J, what is The distance between the charge and the source of the electric field? Round your answer to the nearest tenth.

6 one-sixth parts are needed to make a whole....

A charge of 7.2x10^-5C is placed in an Electric - id721055636785, Kaysofine11icloudcom

bella4269

21.01.2022

d= 2.2 m

Explanation:

The data given in the question is

Charge : q = $7.2 \times 10^{-5} C$

Electric field Strength : E = $4.8 \times 10^{5} N/C$

Electrical potential Energy : U = 75J

Let, the distance be "d"

So, the formula for Electrical potential energy is

$U= q \times E \times d$

Simplified formula for distance become

$d=\frac{U}{q \times E}$

Now , insert the value

$d=\frac{75 J}{7.2 \times 10^{-5}C \times 4.8 \times 10^5 N/C}$

or,

d=2.17 m

Rounding off

d= 2.2 m

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