- #1

pr_angeleyes

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The result of my attempt was:

F= 4kqQb/((d^2/2)+b^2)^3/2

but I dont know if is

F= kqQb/(b^2+l^2/2)^(3/2)

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- Thread starter pr_angeleyes
- Start date

- #1

pr_angeleyes

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The result of my attempt was:

F= 4kqQb/((d^2/2)+b^2)^3/2

but I dont know if is

F= kqQb/(b^2+l^2/2)^(3/2)

- #2

- 8,031

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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude if the total force exerted by the four charges on a charge Q located a distance b along a line perpendicular to the plane of the square and equidistant from the four charges?

The result of my attempt was:

F= 4kqQb/((d^2/2)+b^2)^3/2

The above is almost right. Look again at the term d^2/2. Is it correct?

- #3

Andrew Mason

Science Advisor

Homework Helper

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Perhaps you could explain your reasoning.

The result of my attempt was:

F= 4kqQb/((d^2/2)+b^2)^3/2

but I dont know if is

F= kqQb/(b^2+l^2/2)^(3/2)

What is the l as in l^2/2?

It appears to me that F= 4kqQb/((d^2/2)+b^2)^3/2 is incorrect. Check the d^2/2 term in the denominator. What is the distance of a charge from the centre of the square?

AM

- #4

pr_angeleyes

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F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

=kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

=kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

as in plane parallel to the square net F=0

- #5

pr_angeleyes

- 9

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Perhaps you could explain your reasoning.

What is the l as in l^2/2?

It appears to me that F= 4kqQb/((d^2/2)+b^2)^3/2 is incorrect. Check the d^2/2 term in the denominator. What is the distance of a charge from the centre of the square?

AM

This is what i was thinking:

F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

=kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

=kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

as in plane parallel to the square net F=0

- #6

pr_angeleyes

- 9

- 0

The above is almost right. Look again at the term d^2/2. Is it correct?

this is what i was thinking

F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

=kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

=kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

as in plane parallel to the square net F=0

- #7

- 38,439

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That looks right to me, but I see I'm outvoted byF= 4kqQb/((d^2/2)+b^2)^3/2

- #8

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That looks right to me, but I see I'm outvoted byAMandrude man. Maybe I'm missing something.

How about d^2/2 → (d/2)^2 = d^2/4 ?

- #9

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No, why? As I read the OP, the point Q is distance b from theHow about d^2/2 → (d/2)^2 = d^2/4 ?

- #10

barryj

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I tend to agree with haruspex :-)

- #11

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No, why? As I read the OP, the point Q is distance b from thecentreof the square.

Curses, haruspex and barryj, you are right.

@Andrew Mason:

Locate the four charges at (d/2,d/2,0), (-d/2,d/2,0), -d/2,-d/2,0) and (d/2,-d/2,0) and the observation point at (0,0,b).

Then the distance from any charge is

sqrt[(d/2-0)^2 + (d/2-0)^2 + (0-b)^2] = sqrt{d^2/2 + b^2}.

My excuse: spatial relations was never my strong suit!

- #12

Andrew Mason

Science Advisor

Homework Helper

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You are right. I was using the distance as √2d/2 and it got me a bit confused. Squared it is d^2/2. So I agree the OP's first answer was right. Sorry for any confusion!Curses, haruspex and barryj, you are right.

@Andrew Mason:

Locate the four charges at (d/2,d/2,0), (-d/2,d/2,0), -d/2,-d/2,0) and (d/2,-d/2,0) and the observation point at (0,0,b).

Then the distance from any charge is

sqrt[(d/2-0)^2 + (d/2-0)^2 + (0-b)^2] = sqrt{d^2/2 + b^2}.

My excuse: spatial relations was never my strong suit!

AM

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