A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonconducting 2.50 cm wire. A uniform electric field of magnitude 1.85×108N/C is directed parallel to the wire, as shown in the figure. (Figure 1)
Part A
Find the tension in the wire.
Express your answer with the appropriate units.
Part B
What would the tension be if both charges were negative?
Express your answer with the appropriate units.

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

(a)

The attractive force between the charges is calculated as follows;

[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N[/tex]

The force on the negative charge due to the electric field is calculated as follows;

[tex]F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N[/tex]

The tension on the wire is the resultant of the two forces and it is calculated as follows;

[tex]T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N[/tex]

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N[/tex]

The force on the first negative charge due to the electric field is calculated as follows;

[tex]F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N[/tex]

The force on the second negative charge due to the electric field is calculated as follows;

[tex]F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N[/tex]

The tension on the wire is the resultant of the three forces and it is calculated as follows;

[tex]T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N[/tex]

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