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Misha Larkins [42]
2 years ago
7

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

Physics
1 answer:
Nitella [24]2 years ago
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

#SPJ4

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