A 8.95 μC point charge is glued down on a horizontal frictionless table on the right. If is tied to a -6.35 μC point charge by a light nonconducting wire with length 2.5 cm on the left. The charges are placed in a external, uniform electric field with magnitude 1.65 x 10^8 N/C that is directed parallel to the wire to the right toward the positive charge from the negative charge. Find the tension in the wire. What would the tension by if both charges were negative?

To find the tension in the wire, we need to consider the forces acting on the charges in the system.

Let's first calculate the force experienced by the positive charge due to the external electric field. The force experienced by a charge in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field.

For the positive charge, the force is F1 = (8.95 x 10^-6 C)(1.65 x 10^8 N/C) = 14.7675 N to the right.

Next, let's consider the force between the two charges. The force between two point charges is given by Coulomb's Law, F = k * |q1 * q2| / r^2, where F is the force, q1 and q2 are the charges, r is the distance between the charges, and k is Coulomb's constant (8.99 x 10^9 N * m^2 / C^2).

For the negative charge, the force is F2 = (6.35 x 10^-6 C)(8.95 x 10^-6 C)/(0.025 m)^2 = 13.562 N to the right.

Since the positive charge is to the right, and the negative charge is to the left, these forces are in opposite directions. The net force on the system is the difference between the two forces: F_net = F1 - F2 = 14.7675 N - 13.562 N = 1.2055 N to the right.

Now, we can find the tension in the wire. The tension in the wire is equal to the net force acting on the system. So, the tension in the wire is 1.2055 N.

If both charges were negative, we would follow the same steps as above to calculate the forces. However, since both charges are negative, the forces would be in the opposite direction. The force on the positive charge due to the external electric field would now be to the left, while the force between the charges would still be to the right. Therefore, the net force on the system would be the sum of the forces: F_net = F1 + F2. The tension in the wire would then be equal to the net force, just as in the previous case.

A 8.95 μC point charge is glued down on. "GLUED DOWN" is the key.

Don't yall just want to punch this person so hard that he flies out of the Earth's atmosphere?

You lazy fool. Copying and pasting questions from MasteringPhysics, you should be ashamed of yourself. You will never go anywhere in life.