A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric pote

Question

A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.

What is the electric potential at the center of the sphere?

Express your answer with the appropriate units.

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Delwyn 1 year 2021-09-03T20:51:21+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-09-03T20:52:53+00:00

    Answer:

     
    V_inside = 36 V

    Explanation:

    Given  

    We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

    Required

    We are asked to calculate the potential at the centre of the sphere  

    Solution

    The potential energy due to the sphere is given by equation

    V = (1/4*π*∈o) × (q/r)                                          (1)

    Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

    V ∝ 1/r

    The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

    V_1/V_2=r_2/r_1

    V_inside/V_outside = r/R

    V_inside = (r/R)*V_outside                               (2)

    Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

    V_inside = (1.2 m )/(0.600)*18

                   = 36 V

     
    V_inside = 36 V

    0
    2021-09-03T20:53:11+00:00

    Answer:

    36 V

    Explanation:

    The solid conducting sphere is a positive charge

    and has radius R₁ = 0.6m

    at a point R₂ = 1.20 m, the electric potential V = 18.0 V

    V, electric potential = K q/R where k  = 1/4 πε₀

    V is inversely proportional to R

    V₁ = electric potential at the center

    V₂ = electric potential at 1.2 m

    then

    V₁ /V₂ = R₂ / R₁

    V₁ = V₂ ( R₂ / R₁) = 18.0 V ( 1.2 / 0.6 ) = 36 V

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