대답 1:

Theelectricpotentialduetoapointchargeqatadistance[math]r[/math]is[math]V=14πϵ0(qr).[/math]The electric potential due to a point charge q at a distance [math]r[/math] is [math]V=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}\right).[/math]

Apointchargeof108Cisplacedattheorigin.A point charge of 10^{-8} C is placed at the origin.

ThedistanceofpointAfromtheoriginis[math]rA=42+42+22=6.[/math]The distance of point A from the origin is [math]r_A=\sqrt{4^2+4^2+2^2}=6.[/math]

Theelectricpotentialat[math]A[/math]is[math]VA=14πϵ0(qrA).[/math]\Rightarrow \qquad The electric potential at [math]A[/math] is [math]V_A=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_A}\right).[/math]

ThedistanceofpointBfromtheoriginis[math]rB=12+12+22=3.[/math]The distance of point B from the origin is [math]r_B=\sqrt{1^2+1^2+2^2}=3.[/math]

Theelectricpotentialat[math]B[/math]is[math]VB=14πϵ0(qrB).[/math]\Rightarrow \qquad The electric potential at [math]B[/math] is [math]V_B=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_B}\right).[/math]

Thepotentialdifferencebetweenpoints[math]A[/math]and[math]B[/math]is\Rightarrow \qquad The potential difference between points [math]A[/math] and [math]B[/math] is

VBVA=14πϵ0(qrA)14πϵ0(qrB)\qquad V_B-V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_A}\right)-\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_B}\right)

=14πϵ0q(1rB1rA)=9×109×108×(1319)=20V.\qquad = \frac{1}{4\pi\epsilon_0}\cdot q\left(\frac{1}{r_B}-\frac{1}{r_A}\right) = 9\times 10^9 \times 10^{-8}\times \left(\frac{1}{3}-\frac{1}{9}\right) = 20\,\,V.