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September 15, 2014

September 15, 2014

Posted by **Frustrated** on Friday, February 1, 2013 at 5:22am.

(q = 7.20 x 10^6 C) are fixed at

diagonally opposite corners of a

square with sides of length 0.480 m.

A test charge (q0 = 2.40 x 10^8 C),

with a mass of 6.60 108 kg, is

released from rest at one of the empty corners of the square. Determine

the speed of the test charge when it reaches the center of the square.

- typo -
**Damon**, Friday, February 1, 2013 at 5:36amIf the test charge is ever to approach the center of the square, some charges must be negative and some positive.

In any case the easy way to do this is to compute the potential energy of the test charge at the corners versus at the center. The difference is (1/2) m v^2

- Physics Electricity -
**Devron**, Friday, February 1, 2013 at 6:18amI can not be positive about this answer or solution, but this is what I think.

D=0.480m, so r=0.480m/2

r=0.240m

Q=7.20 x 10^6 C

Qo=2.40 x 10^8 C

Ko=8.99 x10^9 V.m/C

The sum of the potential energies

KE=PE(sum)=Ko(Q/r)+Ko(Q/r) +Ko(Q3/r)

Plug in your values and solve for V

1/2(MV^2)=3Ko(2(Q/r)+(Qo/r)

1/2( 6.60x10^8 Kg)V^2=3(8.99 x10^9 V m/C)[(2(7.20 x 10^6 C/0.240m) +(2.40 x 10^8C/0.240m)]

1/2( 6.60x10^8 Kg)V^2=2.70x10^10V m/C(6 x10^7 C/m +1 x10^9 C/m

1/2(6.60x10^8 Kg)V^2=(2.70x10^10V m/C)(1.06 x 10^7C/M)

1/2(6.60x10^8 Kg)V^2=2.86 x10^19 V

Solving for V,

V=[2*(2.86 x10^19 V/6.60x10^8 Kg)]^1/2

V= 2.94x10^5 m/s

I am not sure if this correct, so be cautious with accepting this as the answer.

- Physics Electricity -
**Devron**, Friday, February 1, 2013 at 7:33amMy initial POST IS INCORRECT and Damon is correct one of the charges has to be negative, but maybe this will help.

r2=0.679m

r=0.480m

Q=7.20 x 10^6 C

Qo=2.40 x 10^8 C

Ko=8.99 x10^9 V.m/C^2

The sum of the potential energies

KE=PE(sum)=[2(Ko(QQo/r)-[(2(KoQQo/r2)

Plug in your values and solve for V

Since I am not sure which are suppose to be negative and which are suppose to be positive, I will just give you the set up and let you solve. I thought the V that I calculated initially was too big, but like I said, I haven't done this in about 8 years.

- Physics Electricity -
**Devron**, Friday, February 1, 2013 at 7:46amAnd divide 0.679/2 to get r2, which is 0.339m

- Physics Electricity -
**Elena**, Friday, February 1, 2013 at 6:10pmPotential of the corner point is

φ=2kq/a.

Potential of the center is

φ₀=2kq2/a√2=2kq√2/a.

Δφ= φ₀ - φ=2kq/a(1-√2)= 2kq 0.41/a.

ΔKE=Work of electric field

mv²/2=q₀Δφ

v=sqrt{ 2q₀Δφ/m) =

=sqrt{2q₀2kq 0.41/am}=

=sqrt{1.64q₀kq/am}

I believe that

q=7.210⁻⁶ C, q₀=2.410⁻⁸ C,

m=6.610⁻⁸ kg,

k =910⁹ Nm²/C²,

a=0.48 m,

v=sqrt {1.64q₀kq /am}=

=sqrt{1.642.410⁻⁸910⁹7.210⁻⁶ /0.48 6.610⁻⁸}=

=283.7 m/s.

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