91影视

As given two charges are at a distance of r=2.50cm and the magnitude of the electric field is $E=1.85脳{10}^{8}N/C$.

Therefore, two forces are exerted on the wire.

The first force is because of a parallel electric field$F_E$

The second force is due to attractive force with the two charges$F_q$

So, the tension (T) between the two charges and as the direction $F_q$is right to left.

Therefore, tension is

role="math" localid="1668179599875" $T=\left|q\right|E-k\frac{q{q}_1}{r^2}$

Here $q=-6.50渭C$is the negative charge and$q_1=+8.75渭C$

Putting all the values:

$$\begin{gathered} T=\left|q\right|E-k\frac{q{q}_1}{r^2} \\ =\left|-6.50脳{10}^{-6}\right|\left(1.85脳{10}^8\right)-\left(9脳{10}^9\right)\frac{\left(-6.50脳{10}^{-6}\right)\left(8.75脳{10}^{-6}\right)}{{\left(0.0250\right)}^2} \\ =382N \end{gathered}$$

Hence, the tension in the wire is 382N.

If both the two charges are negative therefore there will be a repulsive force between them and the direction will be in the same direction as $F_E$.

There the value of T between the two charges is;

$$\begin{gathered} T=F_E+F_a \\ =\left|q\right|E+k\frac{q{q}_1}{r^2} \\ =2020N \end{gathered}$$

Therefore, Tension if both charges were negative 2020N.