d=
qU(
| ||
2mdv02 |
qUL2 |
2m(d+x1)v02 |
得
d2 |
(d+x1)2 |
1 |
4 |
解得:x1=d
(2)斷開(kāi)S后,兩極板間場(chǎng)強(qiáng)E不變,
于是d=
qE(
| ||
2mv02 |
qEL2 |
2mv02 |
得
d |
d+x2 |
1 |
4 |
解得:x2=3d
答:(1)若開(kāi)關(guān)S始終閉合,則這個(gè)距離應(yīng)為d;
(2)若在開(kāi)關(guān)S斷開(kāi)后再移動(dòng)N板,這個(gè)距離又應(yīng)為3d
qU(
| ||
2mdv02 |
qUL2 |
2m(d+x1)v02 |
d2 |
(d+x1)2 |
1 |
4 |
qE(
| ||
2mv02 |
qEL2 |
2mv02 |
d |
d+x2 |
1 |
4 |