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.

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

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.

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.

Well, if these charges were people, this would be quite the electrifying square dance! *ba dum tss*

But let's get serious here. To find the speed of the test charge at the center of the square, we can use conservation of energy.

The potential energy gained by the test charge as it moves closer to the center is equal to the initial potential energy at the empty corner. This can be written as:

q0 * V_initial = q0 * V_final + (1/2) * m * v^2

where q0 is the test charge, V_initial is the initial potential at the empty corner, V_final is the potential at the center of the square, m is the mass of the test charge, and v is its speed at the center.

Now, the potential difference between two points due to point charges is given by:

V = k * |Q| / r

where k is the electrostatic constant (k = 8.99 * 10^9 Nm^2/C^2), |Q| is the magnitude of the charge, and r is the separation between the charges.

Since the charges at the corners are identical, the potential at the center due to each charge would be the same, and its magnitude would be:

V = k * |q| / a, where a is the side length of the square.

With that in mind, you can substitute these values into the conservation of energy equation and solve for v. Just be mindful of the signs when dealing with the charges.

But be careful not to get too shocked by the solution! *wink*

To determine the speed of the test charge when it reaches the center of the square, we can use the principle of conservation of energy.

First, let's calculate the potential energy at the starting position of the test charge. The potential energy between two point charges can be found using the formula:

PE = k * (|q1 * q2| / r)

where PE is the potential energy, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

In this case, the test charge q0 experiences electrostatic potential energy due to the two fixed point charges at the diagonally opposite corners of the square.

PE_start = k * (|q0 * q| / r)

Since the test charge is released from rest, it converts all of its potential energy at the starting position to kinetic energy when it reaches the center of the square.

KE = PE_start

Next, we can calculate the kinetic energy of the test charge once it reaches the center of the square. The kinetic energy can be found using the formula:

KE = (1/2) * m * v^2

where KE is the kinetic energy, m is the mass of the test charge, and v is the speed of the test charge.

Equating the initial potential energy (PE_start) to the final kinetic energy (KE), we can solve for the speed (v):

k * (|q0 * q| / r) = (1/2) * m * v^2

Now let's substitute the given values:

q0 = 2.40 x 10^8 C
q = 7.20 x 10^6 C
r = 0.480 m
m = 6.60 x 10^8 kg
k = 8.99 x 10^9 Nm^2/C^2

Plugging these values into the equation, we can solve for v:

v^2 = (2 * k * |q0 * q| / (m * r))

Finally, take the square root of both sides to find the speed (v):

v = sqrt(2 * k * |q0 * q| / (m * r))

Now you can substitute the given values into the equation and calculate the speed of the test charge when it reaches the center of the square.

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.