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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 magnitude1.85×108N/Cis 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

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

Step by step solution

01

Step 1:

As given two charges are at a distance of r=2.50cm and the magnitude of the electric field is E=1.85×108N/C.

Therefore, two forces are exerted on the wire.

The first force is because of a parallel electric fieldFE

The second force is due to attractive force with the two chargesFq

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

Therefore, tension is

role="math" localid="1668179599875" T=qE-kqq1r2

Here q=-6.50μCis the negative charge andq1=+8.75μC

02

Step 2:

Putting all the values:

T=qE-kqq1r2=-6.50×10-61.85×108-9×109-6.50×10-68.75×10-60.02502=382N

Hence, the tension in the wire is 382N.

03

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 FE.

There the value of T between the two charges is;

T=FE+Fa=qE+kqq1r2=2020N

Therefore, Tension if both charges were negative 2020N.

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