Homework problem (1)

I. 1-8 CBACADDC

Second,

9.-2? 0 E0/3； 4? 0E0 / 3

10.-3? / (2? 0); -? / (2? 0); ? / (2? 0); 3? / (2? 0)

1 1.； Point from point o to the center of the notch.

12.Q /？ 0? ; =0,

Third,

13. Take the charge element, and its charge is

dq =？ dl =？ 0Rsind？

The field strength it produces at point O is

3 points

Two components on the x axis and the y axis.

Dex =-Decosse?

dEy=-dEsin？

Sum the components separately = 0.

∴

14. Solution: Set the coordinate system as shown in the figure. Divide the semi-cylindrical surface into many narrow strips. The linear charge density of a narrow strip with a width of D 1

Take? The position of a line, whose field strength at a point on the axis is

As shown in the figure, its two components on the X axis and Y axis are:

dEx=dE sin？ ，dEy=-dE cos？

Integrate each component separately.

field strength

15. solution: take a thin spherical shell with radius r and thickness dr on the spherical surface, and the charge contained in the shell is

A sphere with radius r contains a total charge of

(r≤R)

Think of this sphere as a Gaussian surface. According to the Gaussian theorem, there are

Get，(r≤R)

The direction is radial, A>0, a.

Make a concentric Gaussian sphere with radius r outside the sphere. According to Gauss theorem, there are

Get，(r > R)

The direction is radial, A>0, a.

16. solution: let the net charge contained in the closed surface be q, because only the x component of the field strength is not zero, the flux of the electric field strength on two planes perpendicular to the x axis is not zero. From Gauss Theorem:

-E 1S 1+ E2S2=Q /？ 0 (S 1 = S2 =S) 3 points.

Then q = 0s (e2-e1) = 0sb (x2-x1).

= ? 0ba2(2a-a) =？ 0ba3 = 8.85× 10- 12 C

Homework Problems (2)

I. 1-8 DBCDDACB

Second,

9. 10cm 10。

1 1.q/(40r 2)； 0 ; q/(40R)； Q / (40r2)

12. The unit positive charge moves around any closed path in electrostatic field, and the work of electric field force is equal to zero potential (or conservative force).

Third,

13. Answer: Think of the charge distribution in the problem as an area density of? What is the density of large plane and surface? The X axis is perpendicular to the plane, the coordinate origin O is in the center of the disk, and the field strength generated by the large plane at X is

The field strength of the disc here is

Zazie Hoko

∴ ?

The potential at this point is

14. solution: according to gauss theorem, e = 0 in the cavity, so the cavity of charged sphere is equipotential region, and the potential at each point is u.

Take a thin spherical surface layer with the radius of R→ R+DR from the spherical surface layer. Its cost is

dq =？ 4? r2dr

The potential generated by the thin layer charge in the center of the ball is

The potential generated by the whole charged sphere at the center of the sphere is

Because the cavity is an equipotential region, the potential u at any point in the cavity is

If you calculate the same score according to the definition of potential.

15. solution: let the charge on the inner ball be q, then the electric field intensity between the two balls is

(R 1