Answer:
r = 0.1 m = 10 cm
Explanation:
To find the position in which the magnitude of the electric field is zero you use the following formula for E:
[tex]E=k\frac{q}{r^2}[/tex]
you can formulate the total electric field of both charges in the following way:
[tex]E_T=k\frac{q_1}{r^2}-k\frac{q_2}{(0.3-r)^2}[/tex]
But ET = 0 for r, the you have:
[tex]E_T=0\\\\k\frac{q_1}{r^2}=k\frac{q_2}{(0.3-r)^2}\\\\(0.3-r)^2q_1=q_2r^2\\\\0.09q_1-0.6q_1r-q_1r^2=q_2r^2\\\\(q_2-q_1)r^2+0.6q_1r-0.09q_1=0[/tex]
Thus, you have a quadratic equation. You replace the values of the charges and then you use the quadratic formula to get the roots of the polynomial:
[tex](16*10^{-16}-4*10^{-6}})r^2+(0.6)(4*10^{-6})r-0.09(4*10^{-6})=0\\\\12r^2+2.4r-0.36\\\\r_{1,2}=\frac{-2.4\pm \sqrt{(2.4)^2-4(12)(-0.36)}}{2(12)}\\\\r=0.1m[/tex]
when you has taken the positive root because it has physical meaning.
hence, the distance in which ET = 0 is 0.1m = 10cm from the smaller charge
Answer:
THE POINT O WHERE THE ELECTRIC INTENSITY IS ZERO IS AT A DISTANCE OF 0.14 m FROM POINT CHARGE A.
Explanation:
Let the point charges be written as qA and qB
qA = 4 * 10^-6 C
qB = 16 *10^-6 C
Distance (d) between the charges is 30 cm = 0.3 m
Let the point where the electric intensity is 0 be 0
Assume that the distance 0 of point chareg A is x
then, the distance 0 from point charge B is 0.3 - x
Since the electric field imtensity is zero at point O,
the intensity at point O due to qA = the intensity at point O due to qB
E (qA) = E (qB)
I/ 4πEo q / d^2 = 1 /4πEo qB/d^2
I/ 4πEo cancels out from both equation, then we have;
4 * 10^-6 / x^2 = 16 *10^-6 / (0.3-x)^2
4 / x ^2 = 16 / (0.3-x)^2
4/x^2 = 16 / 0.09 - x^2
Cross multiply, we have;
4 * 0.09 - x^2 = 16 * x^2
0.36 = 16 x^2 + x^2
0.36 = 17 x^2
x^2 = 0.36 / 17
x^2 =0.02
x = square root of 0.02
x= 0.14 m
So therefore, the point O is at a distance of 0.14 m from point charge qA.
