Answer
The magnitude of the electric field is $~~39~N/C$
Work Step by Step
We can find $E$:
$E = -\frac{dV}{dx}$
$E = (-3000~x)~N/C$
$E = (-3000)~(0.013)~N/C$
$E = -39~N/C$
The magnitude of the electric field is $~~39~N/C$.
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 24 - Electric Potential - Problems - Page 712: 36a
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