Question

A +8.75 $\mathit{\mu }$C point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 $\mathit{\mu }$C point charge by a light, nonconducting 2.50-cm wire. A uniform electric field of magnitude$\mathbf{1}\mathbf{.}\mathbf{85}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{N}\mathbf{/}\mathbf{C}$is directed parallel to the wire, as shown in Fig. E21.34. (a) Find the tension in the wire. (b) What would the tension be if both charges were negative?

Short answer

1. Tension in the wire is 382 N
2. Tension if both charges were negative 2020 N

Step-by-step solution

Step 1:

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

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 $q=-6.50\mu C$is the negative charge and$q_1=+8.75\mu C$

Step 2:

Putting all the values:

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

Hence, the tension in the wire is 382N.

Step 3:

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.