A metal sphere with radius ra = 1.10 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.70 cm . Charge +q is put on the inner sphere and charge −q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 420 V , with the inner sphere at higher potential. Part A Calculate q. qq = Part B Are the electric field lines and equipotential surfaces mutually perpendicular?

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aristocles aristocles · Answer 1 · 2019-10-03T03:10:25+02:00

Answer:

Part a)

[tex]q = 0.58 \times 10^{-9} C[/tex]

Part b)

Yes, equipotential surface is always perpendicular to electric field lines

Explanation:

Potential of inner sphere is given as

[tex]V_{in} = \frac{kq}{r_a} - \frac{kq}{r_b}[/tex]

Potential of outer sphere is given as

[tex]V_{out} = 0[/tex]

Now we know that

[tex]V_{in} - V_{out} = 420 Volts[/tex]

[tex]kq(\frac{1}{r_a} - \frac{1}{r_b}) = 420 V[/tex]

[tex](9\times 10^9)(q)(\frac{1}{0.011} - \frac{1}{0.097}) = 420[/tex]

Now by solving above equation for charge "q"

[tex]q = 0.58 \times 10^{-9} C[/tex]

Part b)

As we know that

[tex]\Delta V = -\int E.dr[/tex]

so here if we need to find the condition of equipotential surface

then we can say

[tex]\Delta V = 0[/tex]

so electric field must be perpendicular to the surface