# (物理)電磁學問題

1.

two small objects,A and B,are fixed in place and separated by 2.00cm.

Object A has a charge of +1.00uC,and object B has a charge of

-1.00uC.How many electrons must be removed from A and put onto

B to make the electrostatic force that acts on each object an attractive

force whose magnitude is 45.0N?

(A) 2.6*10^11 (B) 2.6*10^12 (C) 2.6*10^13 (D) 2.6*10^14

2.

A rectangle has a length of 2d and a height of d.Each of the following

three charges is located at a corner of the rectangle: + q1 (upper left

coener ), + q2 (lower right corner),and - q (lower left corner).

The net electric field at the (empty) upper right corner is zero.Find the

magnitude of q1 and q2.

(A) q1=0.895q , q2= 0.0716q (B) q1=0.0716q , q2=0.895q

(C) q1=0.0895q , q2=0.716q (D) q1=0.716q , q2=0.0895q

### 1 個解答

• 最佳解答

1. 設有x μC的電子從A轉移至B，則A帶電量(1+x) μC，B帶電量 -(1+x) μC，

AB間作用力F=kQq/r^2=9*10^9(1+x)^2*10^-12/0.02^2=45→ x=√2-1

1 mole電子帶電量96500 coul→ (√2-1)μC=(0.414*10^-6/96500)*6.02*10^23=2.6*10^12…(B)

2. +q1對右上角的電場為 kq1/(2d)^2=(1/4)kq1/d^2方向向右

+q2對右上角的電場為 kq2/d^2方向向上

- q對右上角的電場為 (1/5)kq/d^2，方向為右上角指向左下角，故其向左分量=(2/√5)(1/5)kq/d^2=(2/5√5)kq/d^2；向下分量=(1/√5)(1/5)kq/d^2=(1/5√5)kq/d^2

(1/4)kq1/d^2=(2/5√5)kq/d^2；kq2/d^2=(1/5√5)kq/d^2

→ q1=(8/5√5)q=0.716q；q2=(1/5√5)q=0.0894q …(D)

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