Question

一些大学物理题望解答 谢谢 （要过程 完成后追加悬赏） 是electric charge, field, potential相关问题

1. A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 Nm. When the dipole reaches equilibrium its potential energy is equal to -0.2J. What was the inital angle between the direction of the dipole moment and the direction of the electric field?
A.10°
B.90°
C.45°
D.30°
E.0°

2.Two electric chages Qa= +1.0 miuC and Qb= -2.0 miuC are located 0.50m apart. How much work is needed to move the charges apart and double the distance between them?
A. -36*10^(-3)J
B. +18*10^(-3)J
C.0
D.+36*10^(-3)J
E.-18*10^(-3)J

3.The electric potential of a charge distribution is given by V(x,y)= 2xy-x^2-y. At which point is the electric field equal to zero?
A. x=-0.5, y=1
B. x=1, y=1
C. x=1, y=0.5
D. x=0.5, y=0.5
E. x=0, y=0

Answer 1

1，匀强电场中的电偶极子， 施加到其上的最大力矩是0.1 Nm ， 受力平衡时电势能是 -0.2 J 。求电偶极子和电场强度的初始夹角。

2，两个电荷， Qa= +1.0 微库伦， Qb= -2.0 微库伦，相距0.50m 。要使他们之间的距离拉大到初始距离的两倍，需要多少能量？

3，一种电荷分布的电视能由公式 V(x,y)= 2xy-x^2-y 给出。找出电场强度为零的点。

追问

不是要翻译啦。。

追答

Answer 2

3、V（x，y）分别对x，y求偏微分得到得到Ex=2y-2x,Ey=2x-1；电场等于0即Ex=0，Ey=0；所以x=0.5，y=0.5 ，D
2、两电荷看做一个系统，无穷远处的势能看做零，则之间的电势能E = -kQ1Q2/r;做功前系统电势能E1=-kQ1Q2/r，做功后系统电势能E2=-kQ1Q2/2r；需做功W=-（E2-E1）=-kQ1Q2(1/r-1/2r);带入数据计算可得 B 答案
1、这种题 以前没见过，感觉应该数值上满足 sinA = M/E;M为力矩，E电势能，A为初始角；|A|=30 ，D

Answer 3

第三题,E=-gradU,秒了。。 第二题,固定一个电荷 算另外一个的静电势能，算初末互能,秒了。。 第一题,W=qEl,M=qEl*sin(theta),秒了。。