A thin spherical shell with radius R1=3.00R_1 = 3.00 cm is concentric with a larger thin spherical shell with radius R2=5.00R_2 = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge q1=+6.00q_1 = +6.00 nC distributed uniformly over its surface, and the larger shell has charge q2=−9.00q_2 = -9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. What is the electric potential due to the two shells at the following distance from their common center: (i) r=0 r=0; (ii) r=4.00r=4.00 cm; (iii) r=6.00r=6.00 cm?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

문제 25a

23장, 문제 25a

A thin spherical shell with radius $R sub 1 equals 3.00$ cm is concentric with a larger thin spherical shell with radius $R sub 2 equals 5.00$ cm. Both shells are made of insulating material. The smaller shell has charge $q sub 1 equals positive 6.00$ nC distributed uniformly over its surface, and the larger shell has charge $q_{2} = - 9.00$ nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. What is the electric potential due to the two shells at the following distance from their common center: (i) $r equals 0$ ; (ii) $r equals 4.00$ cm; (iii) $r equals 6.00$ cm?

검증된 단계별 안내

Understand that the electric potential due to a spherical shell at a point outside the shell is equivalent to the potential due to a point charge located at the center of the shell. The potential inside a shell is constant and equal to the potential at the surface.

For part (i), at r=0, both shells contribute to the potential. Use the formula for electric potential due to a point charge: V = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q}{r}$. Since r=0, consider the potential at the center due to each shell separately.

For part (ii), at r=4.00 cm, this point is outside the smaller shell but inside the larger shell. Calculate the potential due to the smaller shell using V = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1}{r}$. The potential due to the larger shell is constant inside it and equal to the potential at its surface.

For part (iii), at r=6.00 cm, this point is outside both shells. Calculate the potential due to each shell using V = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q}{r}$ for both q_1 and q_2, and sum them to find the total potential.

Remember to consider the sign of each charge when calculating the potential, as the potential due to a negative charge will be negative. Sum the potentials from both shells to find the total potential at each specified distance.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

13m

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electric Potential

Electric potential is the work done per unit charge in bringing a positive test charge from infinity to a point in space. It is a scalar quantity measured in volts and is related to the electric field. For a point outside a charged spherical shell, the potential is equivalent to that of a point charge located at the center of the shell.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. For spherical symmetry, it simplifies the calculation of electric fields and potentials. Inside a uniformly charged shell, the electric field is zero, affecting the potential calculation at points within the shell.

Superposition Principle

The superposition principle states that the total electric potential at a point is the algebraic sum of the potentials due to individual charges. This principle is crucial for calculating the potential at various distances from the center, as it allows the addition of potentials from both shells, considering their respective charges and distances.