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
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.
5 answers
I 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.
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.
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.
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.
And divide 0.679/2 to get r2, which is 0.339m
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.
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.
1 answer