# Physics Electricity

posted by on .

Two identical point charges
(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  108 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 - ,

If 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 - ,

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 - ,

My 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 - ,

And divide 0.679/2 to get r2, which is 0.339m

• Physics Electricity - ,

Potential of the corner point is
φ=2•k•q/a.
Potential of the center is
φ₀=2•k•q•2/a√2=2•k•q•√2/a.
Δφ= φ₀ - φ=2•k•q/a(1-√2)= 2•k•q• 0.41/a.

ΔKE=Work of electric field
mv²/2=q₀Δφ
v=sqrt{ 2q₀Δφ/m) =
=sqrt{2q₀•2•k•q• 0.41/a•m}=
=sqrt{1.64q₀•k•q/a•m}

I believe that
q=7.2•10⁻⁶ C, q₀=2.4•10⁻⁸ C,
m=6.6•10⁻⁸ kg,
k =9•10⁹ N•m²/C²,
a=0.48 m,

v=sqrt {1.64q₀•k•q• /a•m}=
=sqrt{1.64•2.4•10⁻⁸•9•10⁹•7.2•10⁻⁶ /0.48• 6.6•10⁻⁸}=
=283.7 m/s.