A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x axis. For what finite value(s) of x is (a) the electric field zero? (b) the electric potential zero?
Well, when it comes to electric fields and potentials, things can get a bit shocking! But don't worry, I'm here to brighten things up.
(a) To find the value(s) of x where the electric field is zero, we need to consider the properties of these charges. The charge +q creates an electric field that points away from it, while the charge -2q creates an electric field that points towards it.
Now, let's do some math! Since the electric fields due to each charge point in opposite directions, there must be a point where they cancel each other out, resulting in a net electric field of zero. This can only happen if the magnitudes of the electric fields due to each charge are equal.
So for (a), we need to equate the magnitudes of the electric fields generated by the +q and -2q charges. The formula for the electric field due to a point charge is:
E = k * (|q| / r²)
Since we want the magnitudes to be equal, we can set:
k * (|+q| / r₁²) = k * (|-2q| / r₂²)
Simplifying, we get:
r₁² = 4 * r₂²
r₁ = 2 * r₂
Since r₂ is a finite value at x = 2.00 m, we can conclude that r₁ will be a finite value as well. Therefore, there is a finite value of x at which the electric field is zero.
(b) Now, let's move on to the electric potential! The electric potential is zero when the work done to move a unit charge from infinity to that point is zero. In other words, we need to find the point(s) where the electric potential due to the +q and -2q charges cancel each other out.
Calculating the electric potential due to each charge is a bit trickier, but we can approximate it using the relation:
V = k * (|q| / r)
Similarly to part (a), we need the magnitudes of the electric potentials to be equal. Setting them equal to each other and solving, we would get:
r₁ = 4 * r₂
Again, since r₂ is a finite value at x = 2.00 m, we can conclude that r₁ will be finite as well. Therefore, there is a finite value of x at which the electric potential is zero.
So there you have it! For both the electric field and the electric potential, there is a finite value of x where they are zero. Just remember to keep your charges and puns positive while calculating!
To determine the finite value(s) of x where the electric field is zero, we can use the principle that the electric field at a certain point is the vector sum of the electric fields produced by each individual charge.
(a) To find the finite value(s) of x where the electric field is zero, we need to balance the electric fields produced by the charges +q and -2q.
From the principle of superposition, the electric field at any point P due to charge q at the origin is given by:
E1 = k * q / r^2
where k is the electrostatic constant and r is the distance from the charge q to the point P.
Similarly, the electric field at point P due to charge -2q at x = 2.00 m is given by:
E2 = k * (-2q) / (x - 2)^2
To find the finite value(s) of x where the electric field is zero, we need to find the value of x where E1 = -E2.
Therefore, we have:
k * q / r^2 = k * (-2q) / (x - 2)^2
Simplifying the equation, we get:
q / r^2 = -2q / (x - 2)^2
Cross-multiplying:
(x - 2)^2 = -2r^2
Taking the square root of both sides:
x - 2 = ±√(-2r^2)
Since the square root of a negative number is not a real number, there are no finite values of x where the electric field is zero in this scenario.
(b) To find the finite value(s) of x where the electric potential is zero, we can calculate the electric potential due to each individual charge and set the sum of these potentials to zero.
The electric potential at any point P due to a charge q at the origin is given by:
V1 = k * q / r
where k is the electrostatic constant and r is the distance from the charge q to the point P.
Similarly, the electric potential at point P due to a charge -2q at x = 2.00 m is given by:
V2 = k * (-2q) / (x - 2)
To find the finite value(s) of x where the electric potential is zero, we need to find the value of x where V1 + V2 = 0.
Therefore, we have:
(k * q / r) + (k * (-2q) / (x - 2)) = 0
Simplifying the equation, we get:
q / r = 2q / (x - 2)
Cross-multiplying:
x - 2 = 2r
Therefore, the finite value(s) of x where the electric potential is zero is x = 2r + 2.
To find the finite values of x for which the electric field and electric potential are zero, we need to consider the contributions from both charges and use the principles of superposition.
(a) Finding the finite values of x where the electric field is zero:
The electric field at a point due to a single charge is given by Coulomb's Law:
E = k * (|q| / r²)
where E is the electric field, k is the Coulomb's constant, q is the charge, and r is the distance from the charge.
Considering the charge +q at the origin and the charge -2q at x = 2.00 m, we need to find the point(s) where the sum of the electric fields due to both charges is zero.
Let's consider two points, one on either side of the charge -2q:
1. Point A: To the left of -2q at x < 2.00 m
2. Point B: To the right of -2q at x > 2.00 m
At point A, the electric field due to the charge +q will be directed to the right and the electric field due to the charge -2q will be directed to the left. These fields will have different magnitudes, so the net electric field will not be zero at point A.
Similarly, at point B, the electric fields due to both charges will be in the same direction (to the left), but they will have different magnitudes. Thus, the net electric field will also not be zero at point B.
From this analysis, we can conclude that there is no finite value of x for which the electric field is zero.
(b) Finding the finite values of x where the electric potential is zero:
The electric potential due to a point charge is given by:
V = k * (|q| / r)
where V is the electric potential, k is the Coulomb's constant, q is the charge, and r is the distance from the charge.
To find where the electric potential is zero, we need to find the point(s) where the sum of the potentials due to both charges is zero.
Let's consider two points, one on either side of the charge -2q:
1. Point C: To the left of -2q at x < 2.00 m
2. Point D: To the right of -2q at x > 2.00 m
At point C, the electric potential due to the charge +q will be positive and the electric potential due to the charge -2q will be negative. These potentials will have different magnitudes, so the net electric potential will not be zero at point C.
Similarly, at point D, the electric potentials due to both charges will be negative and have different magnitudes. Thus, the net electric potential will also not be zero at point D.
From this analysis, we can conclude that there is no finite value of x for which the electric potential is zero.
In summary, there are no finite values of x for which the electric field or electric potential is zero in this scenario.
Let the point where E=0 be to the left from the charge q and separated by distance “x” from q and by distance (2+x) from -2q. Then
kq/x²=k2q/(2+x)²,
x²-4x-4 = 0
x= 2±√(4-4) =2.
The point where E=0 is x= - 2m.
Let the point where φ=0 be to the right from q and separated by distance x from it.
kq/x =k2q/(2-x),
x=2/3 m
The point where φ =0 is x=+ 2/3 m.