Question

При перемещении точечного заряда 10 нКл из бесконечности в точку, находящуюся на расстоянии 20 см от поверхности металлического шара, необходимо совершить работу 0.5 мкДж. Радиус шара 4 см. Чему равен потенциал на поверхности шара? PS: мой ответ с правильным отличается знаком - (на этом моменте поподробнее пожалуйста)

Answer 1

q=10^-8 Кл     L=0.2 м    A=5*10^-7 Дж    R=0.04 м    φo=?

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Потенциал шара

φo=k*qo/R    ---->    qo=φo*R/k

Потенциал в исходной точке

φ=k*qo/(L+R)       (потенциал в бесконечности равен 0)

φ=φo*R/(L+R)

Работа по перемещению заряда положительна (потенциальная энергия заряда q в бесконечности равна нулю)

A=q*(φ-0)=q*φo*R/(L+R)   отсюда

φo=(A/q)*(1+L/R)=(5*10^-7/10^-8)*(1+0.2/0.04)=300 B

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Answer 2

Calculation of Potential on the Surface of a Metal Sphere

To calculate the potential on the surface of a metal sphere, we can use the formula:

V = k * Q / r

Where: - V is the potential on the surface of the sphere, - k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2), - Q is the charge of the sphere, and - r is the radius of the sphere.

In this case, we are given the following information: - Charge of the sphere (Q) = 10 nC = 10 x 10^-9 C - Radius of the sphere (r) = 4 cm = 4 x 10^-2 m - Work done to move the charge from infinity to a point 20 cm away from the surface of the sphere = 0.5 μJ = 0.5 x 10^-6 J

To find the potential on the surface of the sphere, we need to calculate the charge of the sphere first. We can use the work-energy principle, which states that the work done on an object is equal to the change in its potential energy.

The work done to move the charge from infinity to a point 20 cm away from the surface of the sphere is equal to the change in potential energy of the charge. Therefore, we can write:

Work done = Change in potential energy

0.5 x 10^-6 J = Q * V

Since the charge (Q) is given as 10 nC, we can substitute the values into the equation:

0.5 x 10^-6 J = (10 x 10^-9 C) * V

Simplifying the equation, we find:

V = (0.5 x 10^-6 J) / (10 x 10^-9 C)

V = 50 V

Therefore, the potential on the surface of the metal sphere is 50 volts.

Please note that the sign of the potential depends on the sign of the charge. In this case, since the charge is positive, the potential is positive as well.

I hope this helps! Let me know if you have any further questions.