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RSB [31]
3 years ago
8

At what distance r from a point charge is the electric potential due to that point charge 0 v?

Physics
1 answer:
galben [10]3 years ago
8 0

Answer:

Infinite Distance

Explanation:

The electric potential due to a point charge can be expressed by the following equation:

V=\frac{kQ}{r}

Here,

V is the electric potential due to the point charge

k is the proportionality constant

Q is the magnitude of the point charge

r is the distance from the charge

As the value of r increases, the value of V decreases since there is an inverse relation between the two. The value of V can be absolutely 0 when the distance from the charge is infinite i.e. r is infinite. Mathematically, dividing a number by infinity results in zero. Also theoretically speaking, at infinite distance the electric field lines won't approach and hence the electric potential would be zero.

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Answer it fast and no spamming....! !
ExtremeBDS [4]

Answer:

When the resistances are connected in the parallel the equivalent resistance will be always less than the value of the lowest resistance in the parallel circuit.

The equivalent resistance in the parallel circuit can be calculated using the following formula.

1/R=1/R1+1/R2+1/R3

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3 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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