Question
Please answer parts A,B,&C. Thank you!
p06 infinity.) Part C
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Answer #1

Given is:-

The potential  V4.68volts

and electric field  E = 15V/m

Now,

part-a

We know that the electric field for any charge is given by

E = rac{V}{d}

by plugging all the values we get

15 = 4.00

which gives us

d-0.312772

Part-b

The potential due to the charge is given by

KQ

by plugging all the values we get

(9 x 10)Q 4.680.312

which gives us

Q = 1.62 × 10-10C

Part-c

As,the electric potential at the given point is positive, then the given charge is positive.

Hence,

The direction of electric field is away    from the given positive point charge

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Please answer parts A,B,&C. Thank you! p06 infinity.) Part C
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