AL phy 2008年paper2A 第34題

Q: 根據公式應該是.....E = kq/r^2
k is consistent
那現在q是負數
答案不是應該C (即2 is correct)嗎?

Since the point charge is -ve, the equation for electric field becomes,
E = -kq/r^2
Bearing in mind that the -ve sign only gives the direction of electric field (i.e. the field lines are pointing towards the charge). The magnitude of the field is given by kq/r^2.

It is then apparent that the larger the value of r, the lower is the value for E. Hence, electric field at R (with larger value of r as compared to that at P) is weaker than the field at P.

(1) 記住 E=-dV/dx。電場是由高電勢指向低電勢﹐因此電場是由R指向P

(2) 由圖知電荷放在P的左邊﹐又電場的變化是連續的﹐既然R距離電荷較P遠﹐所以R的電場應該弱過P﹐因此敘述錯誤。

(3) 電勢與距離成反比。因此從P點開始﹐電勢開始時是增加得比較快﹐其後漸漸減慢﹐因此R點的電勢應該已經高過 -40V。

ANSWER: A

Yes, E-field is a vector. But bear in mind that "E-field stronger" refers to "greater magnitude", that is, similar to concept in absolute values like:

|-5| > |-3| although -5 < -3

So for a point negative charge (we can deduce that the charge is negative by the fact that the E-field lines are pointing inwards since the equipotential line of -30V is outer than -50V according to the diagram):

By the equation V = kq/r and E = kq/r2

-50 = kq/r1 and -30 = kq/r2

|kq/r1| > |kq/r2|

|1/r1| > |1/r2|

r1 < r2

The -50 V potential should be closer to the charge than the -30 V potential

Therefore:

E1 = kq/r12 and E2 = kq/r22.

|kq/r12| > |kq/r22| since r1 < r2

|E1| > |E2|

Hence the E-field should be stronger at P than at R