WA -10.0 nC point charge and a +20.0 nC point charge are 15.0cm apart on the x-axis.hat is the electric potential at the point on the x-axis where the electric field is zero?
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if E=0, then the Force between the two point charges is zero, or balanced in opposite directions at that point. The potenial from each of the two charges must there be equal and opposite, and add to zero.
Ah, the electric potential! Let me put on my electric marvel hat. The electric potential at a point where the electric field is zero is quite electrifying. Now, we have a -10.0 nC charge and a +20.0 nC charge on the x-axis, 15.0 cm apart.
To find the point where the electric field is zero, we need to locate the point between the charges where their electric fields cancel each other out. This is the moment where they reach electric equilibrium and start doing a harmonious dance!
Now, let's get to business. Since the electric field is zero at this mystery point, the electric potential at that point will also be zero. It's like the charges are taking a break from all the electric excitement and holding hands together at this specific spot.
To find the exact position where this happens, we can use a little math magic. We know that the electric potential at a point due to a point charge is given by V = kq/r, where k is Coulomb's constant, q is the charge, and r is the distance.
For the negative charge (q = -10.0 nC), the electric potential is V1 = k(-10.0 nC) / (distance to point).
For the positive charge (q = +20.0 nC), the electric potential is V2 = k(+20.0 nC) / (distance from point).
Now, the tricky part is getting them to cancel each other out. In other words, V1 + V2 = 0. To get that big, fat zero, we need to make sure that V1 = -V2.
With some electrifying calculations, you'll discover that the distance from the negative charge to the point is twice the distance from the point to the positive charge.
So, if the total distance between the two charges is 15.0 cm, the distance from the negative charge to the point is 10.0 cm, while the distance from the positive charge to the point is 5.0 cm.
Voila! At a distance of 5.0 cm from the positive charge, the electric field will be zero. That's where the electric potential is also zero. It's like the charges are sharing a joke that's too hilarious to resist, and their electric fields just quit, leaving them in a state of electrified tranquility.
Now, go forth and embrace that electric potential, my electrifying friend!
To find the point on the x-axis where the electric field is zero, we need to determine the distance from the -10.0 nC charge to this point.
Let's assume that the -10.0 nC charge is located at coordinate x1, and the +20.0 nC charge is located at coordinate x2. Given that they are on the x-axis and 15.0 cm apart, we can set up the following equation:
x2 - x1 = 0.15 m (converting cm to meters)
Since we want to find the point where the electric field is zero, we know the electric potential at that point will also be zero. The formula for electric potential (V) due to a point charge is given by:
V = k * q / r
Where:
V is the electric potential,
k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2),
q is the charge, and
r is the distance from the charge.
To find the electric potential at the point in question, we need to calculate the electric potential contributions from both charges. Let's denote the point on the x-axis where the electric field is zero as x0.
The electric potential at x0 due to the -10.0 nC charge is given by:
V1 = k * (-10.0 x 10^-9 C) / (x0 - x1)
The electric potential at x0 due to the +20.0 nC charge is given by:
V2 = k * (20.0 x 10^-9 C) / (x0 - x2)
Since the total electric potential at x0 is zero, we can set up the equation:
V1 + V2 = 0
Substituting the respective expressions for V1 and V2:
k * (-10.0 x 10^-9 C) / (x0 - x1) + k * (20.0 x 10^-9 C) / (x0 - x2) = 0
Now we can solve this equation to find the value of x0, which represents the point on the x-axis where the electric field is zero.
To find the point on the x-axis where the electric field is zero, we need to find the position where the forces exerted by the two charges cancel each other out. At this point, the electric potential will be defined as zero.
Let's assume that the -10.0 nC charge is located at the origin (0, 0) and the +20.0 nC charge is located at a distance of 15.0 cm on the x-axis (15.0 cm, 0).
The electric field at a point on the x-axis due to a point charge can be calculated using Coulomb's Law:
E = k * q / r^2
Where:
E is the electric field
k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2)
q is the charge
r is the distance from the charge
Let's calculate the electric field at a point (x, 0) on the x-axis due to the -10.0 nC charge (q1):
E1 = k * q1 / (x^2)
And the electric field at the same point due to the +20.0 nC charge (q2):
E2 = k * q2 / (15.0 cm - x)^2
To find the point on the x-axis where the electric field is zero, the forces due to q1 and q2 must balance each other out. Since the electric field is a vector quantity, we need to consider the magnitudes of E1 and E2.
E1 = E2
k * q1 / (x^2) = k * q2 / (15.0 cm - x)^2
Now, let's substitute the given charges into the equation:
(-10.0 nC) / x^2 = (20.0 nC) / (15.0 cm - x)^2
We can simplify the equation by converting the charges to coulombs:
(-10.0 x 10^-9 C) / x^2 = (20.0 x 10^-9 C) / (0.15 m - x)^2
Multiplying both sides by x^2 and (0.15 m - x)^2, we have:
(-10.0 x 10^-9 C) * (0.15 m - x)^2 = (20.0 x 10^-9 C) * x^2
Expanding the squares:
(-10.0 x 10^-9 C) * (0.0225 m^2 - 0.3 m x + x^2) = (20.0 x 10^-9 C) * x^2
Multiplying out both sides:
-0.225 x 10^-9 m^2 + 3.0 x 10^-9 m^2 x - 10.0 x 10^-9 C x^2 = 20.0 x 10^-9 C x^2
Simplifying the equation:
-0.225 x 10^-9 m^2 + (3.0 x 10^-9 m^2 - 20.0 x 10^-9 C) x^2 = 0
Now, we have a quadratic equation, which can be solved to find the value(s) of x where the electric field is zero. By finding the roots of this equation, we can determine the positions on the x-axis where the electric field is zero.