For a single, isolated point charge carrying a charge of q = 5.22 x 10 - 11 C, o

Question

For a single, isolated point charge carrying a charge of q = 5.22 x 10 - 11 C, one equipotential surface consists of a sphere of radius 0.0154 m centered on the point charge as shown. What is the potential on this surface? You would like to draw an additional equipotential surface, which is separated by 10.2 V from the previously mentioned surface. How far from the point charge should this surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface.

Explanation / Answer

here by using the formula

V1 = k * q / r1

V1 = 9 * 10^9 * 5.22 * 10^-11 / 0.0154

V1 = 30.5 V

then the potential difference between V1 and V2

delta V = V1 - V2 = V1 - k * q / r2

therefore the radius r2 is

r2 = (9 * 10^9 * 5.22 * 10^-11) / ( 30.5 - 10.2 )

r2 = 0.0231 m

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.