Question

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

Answer 1

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.

What are equipotential surfaces?

• 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,

                          $r_d=r_1-r_2$

How to solve the problem?

• 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:

https://brainly.com/question/28044747

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