Q76P A metal sphere with radius R1 ha... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q76P (page 782)

A metal sphere with radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Short Answer

Expert verified

(a) Theelectric field at the surface of the sphere is $\frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}^{2}}$ and electric potential at the surface of the sphere isrole="math" localid="1668309845269" $\frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}^{}}$ .

(b) The charge on the first sphere is $q_{1} = \frac{Q_{1} R_{1}}{R_{1} + R_{2}}$ and on the second sphere isrole="math" localid="1668309885903" $q_{2} = \frac{Q_{1} R_{2}}{R_{1} + R_{2}}$ .

(c) The electric potential at the surface of each sphere is $V = \frac{1}{4 蟺 {\overset{藱}{o}}_{2}} \frac{Q_{1}}{R_{1} + R_{2}}$ .

(d) The electric field at the surface of sphere with radius R₁ is $\frac{1}{4 蟺 o} \frac{Q_{1}}{(R_{1} + R_{2}) R_{1}}$ and sphere with radius R₂ is $\frac{1}{4 蟺 {\overset{藱}{虁}}_{s}} \frac{Q_{1}}{(R_{1} + R_{2}) R_{2}}$ .

Step by step solution

Step 1:

The electric field inside a conducting sphere is zero and at the surface of the sphere is given by

$E = \frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}^{2}}$

The electric potential is constant for the entire sphere as no work is done on the charges. So the potential at the surface depends on the radius of the sphere and it is given by

$V = \frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}}$

Therefore, the electric field at the surface of the sphere is $\frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}^{2}}$ and electric potential at the surface of the sphere is $\frac{1}{4 蟺蔚_{0}} \frac{Q_{1}}{R_{1}}$ .

Step 2:

This sphere is connected by a long thin conducting wire to another sphere of radius R₂ and charge Q₂ till electrostatic equilibrium has been reached.

The electrostatic equilibrium has been reached, this means that the potential at the surface of both spheres is the same and V₁ = V₂.

Also, the charge Q₁ after the contact will be conservative, which means after the contact the charge Q₁ will be divided to charge q₁ on the first sphere and q₂ on the second sphere, so the charge at electrostatic equilibrium will be

$\begin{aligned} Q_{1} = q_{1} + q_{2} \\ q_{1} = Q_{1} 鈭� q_{2} \end{aligned}$

The charge q₁ and q₂ will be

$\begin{aligned} V_{1} = V_{2} \\ \frac{1}{4 蟺 o} \frac{q_{1}}{R_{1}} = \frac{1}{4 蟺 o} \frac{q_{2}}{R_{2}} \\ \frac{q_{1}}{R_{1}} = \frac{q_{2}}{R_{2}} \\ \frac{Q_{1} 鈭� q_{2}}{R_{1}} = \frac{q_{2}}{R_{2}} \\ q_{2} = \frac{Q_{1} R_{2}}{R_{1} + R_{2}} \end{aligned}$

And the charge q1 on the sphere will be

$\begin{aligned} q_{1} = Q_{1} 鈭� \frac{Q_{1} R_{2}}{R_{1} + R_{2}} \\ q_{1} = \frac{Q_{1} R_{1}}{R_{1} + R_{2}} \end{aligned}$

Therefore, the charge on the first sphere is $q_{1} = \frac{Q_{1} R_{1}}{R_{1} + R_{2}}$ and on the second sphere is $q_{2} = \frac{Q_{1} R_{2}}{R_{1} + R_{2}}$ .

Step 3:

The electric potential at the surface of each sphere V₁ and V₂ is given by

$\begin{aligned} V_{1} = \frac{1}{4 蟺 o} \frac{q_{1}}{R_{1}} \\ V_{1} = \frac{1}{4 蟺 o} \frac{(Q_{1} R_{1}) / (R_{1} + R_{2})}{R_{1}} \\ V_{1} = \frac{1}{4 蟺 o} \frac{Q_{1}}{R_{1} + R_{2}} \end{aligned}$

As V1=V2, the potential at each sphere is $V = \frac{1}{4 蟺 o_{2}} \frac{Q_{1}}{R_{1} + R_{2}}$

Step 4:

The electric field at the surface E₁ and E₂ is given by

$\begin{aligned} E_{1} = \frac{1}{4 蟺 a} \frac{q_{1}}{R_{1}^{2}} \\ = \frac{1}{4 蟺 a} \frac{(Q_{1} R_{1}) / (R_{1} + R_{2})}{R_{1}^{2}} \\ = \frac{1}{4 蟺 a} \frac{Q_{1}}{(R_{1} + R_{2}) R_{1}} \\ E_{2} = \frac{1}{4 蟺 a} \frac{q_{2}}{R_{2}^{2}} \\ = \frac{1}{4 蟺 a} \frac{(Q_{1} R_{1}) / (R_{1} + R_{2})}{R_{2}^{2}} \\ = \frac{1}{4 蟺脿} \frac{Q_{1}}{(R_{1} + R_{2}) R_{2}} \end{aligned}$

The electric field at the surface of sphere with radius R₁ is $\frac{1}{4 蟺 o} \frac{Q_{1}}{(R_{1} + R_{2}) R_{1}}$ and sphere with radius R₂ is $\frac{1}{4 蟺 o_{1}} \frac{Q_{1}}{(R_{1} + R_{2}) R_{2}}$ .