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Q29PE

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College Physics (Urone)
Found in: Page 697

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Short Answer

If the potential due to a point charge is 5.00×102 V at a distance of 15.0 m, what are the sign and magnitude of the charge?

The potential V is positive, and the sign of the charge Q is also positive which is obtained as: Q=8.33×10-3C.

See the step by step solution

Step by Step Solution

Step 1: Given Data

Potential due to the charge is 5.00×102 V.

Distance from the charge is 15.0 m.

Step 2: Define Electric Field

A physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attractive or repulsive, is referred to as an electric field.

Step 3: The given values

Finding the magnitude and sign of the charge Q, if the potential is due to a point charge is: V=5.00 x10 V,

It is at a distance of r which is: r = 15.0 m.

Step 4: Obtaining the value of :

In the equation of the potential V is: V=kQr

Determine the charge Q, as: Q=rVk

The value of k= k=9.00×109N.m2C2.

Step 5: Evaluation of the charge

Substituting the quantities in the equation for the charge Q as:

Q=rVk=(15.0 m)(5.00×102V)9.00×109 N.m2C2=8.33×10-3 C

Therefore, the potential V is positive and the sign of the charge Q is positive which gives us: Q=8.33×10-3C.

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