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What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?

Respuesta :

skyluke89
The potential at a distance r from a charge Q is given by
[tex]V(r) = k_e \frac{Q}{r} [/tex]
where ke is the Coulomb's constant.

The charge in our problem is [tex]Q=2.9 nC=2.9 \cdot 10^{-9} C[/tex]; for the point at [tex]r=0.48 mm=0.48 \cdot 10^{-3} m[/tex], the potential is
[tex]V_1 = k_e \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}= 5.43 \cdot 10^4 V[/tex]

For the point at infinity, we immediately see that the potential is zero, because [tex]r= \infty [/tex] and so [tex]V_2 = 0[/tex].

Therefore, the potential difference between the two points is
[tex]\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V[/tex]
okpalawalter8

The potential difference is mathematically given as

[tex]dV= 5.43*10^4 V[/tex]

What is the potential difference between the two points?

Question Parameters:

the potential difference between a point 0.48 mm

from a charge of 2.9 nc and a point at infinity

Generally, the equation for the   is mathematically given as

[tex]V_1 = k_e \frac{Q}{r}[/tex]

[tex]V_1 = (8.99 * 10^9 Nm^2 C^{-2}) \frac{2.9 * 10^{-9} C}{0.48* 10^{-3} m}[/tex]

V_1=  5.43 *10^4 V

Therefore, the potential difference is

d V = V_1 - V_2

[tex]dV= 5.43*10^4 V[/tex]

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