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Anonymous asked in Science & MathematicsPhysics · 2 months ago

what is the potential on this surface? ?

For a single, isolated point charge carrying a charge of pC, q=57.7 pC, one equipotential surface consists of a sphere of radius r1=29.4 mm centered on the point charge.  

 

What is the potential on this surface?

To draw an additional equipotential surface separated by V 11.7 V from the previous surface, how far from the point charge should this second surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface is. 

1 Answer

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  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    potential = kQ / r

    where k = 1/4πε₀ = 8.99e9N·m²/C²

    so

    potential = 8.99e9N·m²/C² * 57.7e-12C / 0.0294m

    potential = 17.6 V ◄

    then

    11.7 V = 8.99e9N·m²/C² * 57.7e-12C / r

    solves to

    r = 0.0443 m = 44.3 mm ◄

    If you find this helpful, please select Favorite Answer!

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