Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sph

Question

Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?

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Thu Cúc 5 years 2021-08-22T22:19:31+00:00 1 Answers 19 views 0

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  1. Neala
    0
    2021-08-22T22:20:50+00:00
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    Answer:

    ΔV = -1321.73V

    Explanation:

    The change in potential along the path from C to D is given by the following expression:

    \Delta V=-\int_a^bE dr         (1)

    E: electric field produced by a charge at a distance of r

    a: distance to the sphere at position C = 0.021m

    b: distance to the sphere at position D = 0.055m

    The electric field is given by:

    E=k\frac{Q}{r^2}                 (2)

    Q: charge of the sphere = 5nC = 5*10^-9C

    k: Coulomb’s constant = 8.98*10^9Nm^2/C^2

    You replace the expression (2) into the equation (1) and solve the integral:

    \Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b            (3)

    You replace the values of a and b:

    \Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V

    The change in the potential along the path C-D is -1321.73V

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