A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m3. What is the magnitude of the potential difference b

Question

A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m3. What is the magnitude of the potential difference between the center and a point 4.0 cm away?

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Xavia 1 month 2021-08-14T06:58:16+00:00 1 Answers 1 views 0

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    2021-08-14T06:59:26+00:00

    Explanation:

    The given data is as follows.

       Radius of the sphere (R) = 10 cm = 0.01 m   (as 1 m = 100 cm)

       Distance from the center (r) = 4 cm = 0.04 m

       Charge density (\sigma) = 100 nC/m^{3}

                                      = 100 \times 10^{-9} C/m^{3}   (As 1 nm = 10^{-9} m)

    As the relation between charge and potential difference is as follows.

              Q = \sigma V

                  = \sigma (\frac{4}{3} \pi r^{3})

                  = 100 \times 10^{-9} C/m^{3} \times \frac{4}{3} \pi (0.01)^{3}  

                  = 4.19 \times 10^{-10} C

    Expression for electric field is as follows.

               E(r) = k \frac{qr}{R^{3}}

    Electric potential, V(r) = -\int_{0}^{r} E(r) dr

                   = -\int_{0}^{r} k \frac{qr}{R^{3}} dr

                   = -k (\frac{qr^{2}}{2R^{3}})  

                   = -9 \times 10^{9} Nm^{2}/C^{2} (\frac{4.19 \times 10^{-10} C (0.04)^{2}}{2(0.1)^{3}})  

                  = -3 V

    Thus, we can conclude that the magnitude of the potential difference between the center and a point 4.0 cm away is -3 V.

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