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tangare [24]
2 years ago
13

An electric charge of 1 microcoulomb is affected by an electric field of intensity 1000 newtons per coulomb at a point, then thi

s point is a meter away from it by a distance
Physics
1 answer:
Alchen [17]2 years ago
8 0

Answer:

r = 3 m

Explanation:

Given that,

Charge, q=1\ mu C

Electric field at a point, E=1000\ N/C

We need to find the distance away from it. We know that electric field at point r is given by :

E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 10^{-6}}{1000}} \\\\r=3\ m

So, the required distance is 3 m.

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We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
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We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
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\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
Now we plug in the second equation into first one:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0
We end up with quadratic equation that we have to solve, I won't solve it by hand. 
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Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
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We do the same thing here, but we must express v_a from momentum equation:
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Solutions are:
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