Unformatted text preview: 9. At points between the charges, the individual electric ﬁelds are in the same direction and do not cancel.
Charge q2 has a greater magnitude than charge q1 , so a point of zero ﬁeld must be closer to q1 than to
q2 . It must be to the right of q1 on the diagram. •
q2 d P
• •
q1 We put the origin at q2 and let x be the coordinate of P , the point where the ﬁeld vanishes. Then, the
total electric ﬁeld at P is given by
E= 1
4πε0 q2
q1
−
2
x
(x − d)2 where q1 and q2 are the magnitudes of the charges. If the ﬁeld is to vanish,
q2
q1
=
.
x2
(x − d)2
√
√
We take the square root of both sides to obtain q2 /x = q1 /(x − d). The solution for x is
√ x=
=
=
=
The point is 50 cm to the right of q1 . q2
d
√
q2 − q1
√
4.0q1
√
d
√
4.0q1 − q1
√ 2.0
d = 2. 0 d
2.0 − 1.0
(2.0)(50 cm) = 100 cm . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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