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Old Aug 29th 2008, 07:31 PM   #1
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Unhappy Pls help....just cant remember, but an easy problem related to ELECTRIC FIELD.

q1 .______________________.q2
0.1meters

The above fig. shows 2 point charges q1 (1x10-6 C) and q2 (2x10-6 C) are 0.1m away. On which point on the joining line is the Electrin Field Steangth zero??
(the smaller '-6's r meant to be the powers of 10)
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Old Aug 29th 2008, 08:12 PM   #2
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Originally Posted by Ess J View Post
q1 .______________________.q2
0.1meters

The above fig. shows 2 point charges q1 (1x10-6 C) and q2 (2x10-6 C) are 0.1m away. On which point on the joining line is the Electrin Field Steangth zero??
(the smaller '-6's r meant to be the powers of 10)
The electric field a distance r from a point particle of charge q is given by
$\displaystyle E = \frac{k q}{r^2}$
or
$\displaystyle E = \frac{1}{4 \pi \epsilon _0} \frac{q}{r^2}$
depending on how your book does it. The field is a vector field and points away from a positive charge and toward a negative charge.

Find the field due to each point charge and add them vectorally. Putting the origin at q1 and the positive x direction toward q2 we then have at a distance x from q1:
$\displaystyle E = \frac{k q_1}{x^2} - \frac{k q_2}{(0.1 - x)^2}$

Set E = 0 and solve for x.

-Dan
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Last edited by topsquark; Aug 31st 2008 at 10:30 AM.
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Old Aug 30th 2008, 08:37 PM   #3
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Thank u so very much...it was my first question that i posted & got it answered so perfectly!! thanks agyn
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Old Aug 31st 2008, 09:50 AM   #4
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well dan,
wen i was tryin to solve:

, i am still stuck in the '0.1-x' part... shuld'nt the

whole thing be squared too? I mean ; (0.1-x)^2.

It'll b a great help...Tc.
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Old Aug 31st 2008, 10:31 AM   #5
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Originally Posted by Ess J View Post
well dan,
wen i was tryin to solve:

, i am still stuck in the '0.1-x' part... shuld'nt the

whole thing be squared too? I mean ; (0.1-x)^2.

It'll b a great help...Tc.
Yes, that was a typo. I have fixed it in the original post and thank you for the catch.

-Dan
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