Answer:
please the answer below
Explanation:
(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is
[tex]V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\[/tex]
k=8.89*10^9
For both cases we have
[tex]k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}[/tex]
(b) by replacing this values of r in the expression for V we obtain
[tex]k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}[/tex]
hope this helps!!
Answer:
a) d = 2.48 cm
b) [tex]\frac{q_1}{q_2}=[/tex] 1.48 cm
Explanation:
The potential at the point [tex]x_1[/tex] = 1.0 cm to the left of the charge is given as:
[tex]V = \frac{kq_1}{d-x_1}-\frac{kq_2}{x_1} \\ \\[/tex]
since V = 0 ; Then:
[tex]0 = \frac{kq_1}{d-x_1}-\frac{kq_2}{x_1} \\ \\\frac{kq_2}{x_1} = \frac{kq_1}{d-x_1}[/tex]
[tex]\frac{q_1}{q_2} = \frac{d -x_1}{x_1}[/tex]
[tex]\frac{q_1}{q_2} = \frac{d }{1.0 \ cm} -1[/tex]
The potential at the point [tex]x_2[/tex] = 5.2 cm to the right of the negative charge is:
[tex]V = \frac{kq_1}{x_2+d}-\frac{kq_2}{x_2}[/tex]
since V = 0
[tex]\frac{kq_2}{x_2} =\frac{kq_1}{x_2+d}[/tex]
[tex]\frac{q_1}{q_2} = \frac{x_2+d}{x_2}[/tex]
[tex]\frac{q_1}{q_2} = 1+ \frac{d}{x_2}[/tex]
[tex]\frac{q_1}{q_2} = 1+ \frac{d}{5.2 \ cm}[/tex]
Now, let's solve for d (the distance between the charges ) from the above derived formulas
If we represent the ratio of [tex]\frac{q_1}{q_2} = a[/tex]
Then; [tex]a = \frac{d}{1.0} -1[/tex]
[tex]a+1 = \frac{d}{1.0}[/tex]
1.0a + 1 = d ------- equation (1)
Also;
[tex]a = 1+ \frac{d}{5.2}[/tex]
[tex]a = \frac{5.2+d}{5.2}[/tex]
5.2 a = 5.2 +d
5.2 a - 5.2 = d ---------- equation (2)
From equation (1) ; lets replace d = 5.2 a - 5.2
then :
1.0 a + 1 = d
1.0 a + 1 = 5.2 a - 5.2
1.0a - 5.2 a = - 5.2 - 1
- 4.2 a = -6.2
a = 1.48 cm
Also replace a = 1.48 cm into equation (1) to solve for d
1.0 a + 1 = d
1.0 (1.48 )+ 1 = d
1.48 + 1 = d
d = 2.48 cm
b)
The ratio of the magnitude of the charges :
[tex]\frac{q_1}{q_2}= \frac{d}{1.0} -1[/tex]
[tex]\frac{q_1}{q_2}= \frac{2.48}{1.0} -1[/tex]
[tex]\frac{q_1}{q_2}= \frac{1.48}{1.0}[/tex]
[tex]\frac{q_1}{q_2}=[/tex] 1.48 cm
