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2 years ago

What is the radial distance between the 500 v equipotential surface and the 1000 v surface?.

1 answer:

6 0

Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.

Explanation: To find the answer, we have to know more about the equipotential surfaces.

<h3>What are equipotential surfaces?</h3>

An equipotential surface is the locus of all points which have the same potential due to the charge distribution.
Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.
We have the equation for electric potential as,

$V=\frac{Qk}{r}$ , where k is equal to 1/(4π∈₀) = $8.99*10^9$ .

equation for radial distance will be,

$r_d=r_1-r_2$

<h3>How to solve the problem?</h3>

For the first surface, we can write the equation of potential as,

$500V=\frac{Qk}{r_1} \\thus, r_1=\frac{Qk}{500V}= (1.79*10^7 *Q) m$

For the second surface, we can write the equation of potential as,

$1000V=\frac{Qk}{r_2}\\ r_2= \frac{Qk}{1000V} =(8.99*10^6*Q) m$

Thus, the radial distance will be,

$r_d=1.79*10^7Q-8.99*10^6Q=(8.91*10^6*Q)m$

Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.

Learn more about the equipotential surface here:

brainly.com/question/28044747

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