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by：Tianwang
2020-06-24

Elite tutors network & gt;
High school physics & gt;
If removed AB, the rest with electric charge distribution is changeless, point P potential into c, now let CD rod D end clockwise rotate 90 °, O, P point of electric potential at this time, phi phi O P, respectively, of the following statements is correct (
)
答:φO = $ 压裂{
1}
{
2}
c + 压裂{
3}
{
8}
a$ B． φP=c C． φO=$frac{
1}
{
2}
c -美元
$frac{
1}
{
8}
d . φP = c + 美元压裂{
1}
{
4}
So CD into the CD 'didn't change the field strength point P, still to c for O points, four bonzi completely symmetrical, so each rod produced by the electric potential is $ frac {
1}
{
4}
$a, AD, BC did not change the position, so produce potential $ frac {
1}
{
2}
一美元;
CD 'produce potential analysis as follows: see the AD, BC to P potential, CD to P $ frac {
1}
{
4}
$a, AD, BC, CD for total c, so the AD, BC to c -
$frac{
1}
{
4}
一美元;
Two bonzi is symmetrical, so any potential $ frac {
1}
{
2}
$ (
c -
$frac{
1}
{
4}
一个美元)
, for this is the BC P potential;
Looking at the CD 'potential of O, think of the electrostatic field distribution, can know the potential root BC to P;
Potential superposition to point O: phi O = $ frac {
1}
{
2}
c + 压裂{
3}
{
8}
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