A charge +q is at the origin. A charge -2q is at x = 6.30 m on the +x axis.
(a) For what finite value of x is the electric field zero?
(b)or what finite values of x is the electric potential zero? (Note: Assume a reference level of potential V = 0 at r = ∞.)
Smallest value of x:
Largest value of x:
I m not sure really where to start with this.. any guidance would be appreciated.
a. add the E from each charge, set to zero.
k(q)/x^2+-2kq/(6-x)^2=0
(6-x)^2-2x^2=0
solve that quadratic for x.
b. Voltage=kq/x
kq/x-2kq/(6-x)=0
6-x-2x=0
x=2
To approach this problem, we need to use the concepts of electric field and electric potential due to a point charge. Let's break it down step by step:
(a) For what finite value of x is the electric field zero?
Step 1: Determine the expression for the electric field at any point on the +x axis due to the charges.
The electric field at a point due to a point charge q is given by Coulomb's law as:
E = k*q/r^2
where E is the electric field, q is the charge, r is the distance from the charge, and k is the electrostatic constant.
In this case, the electric field at any point on the +x axis due to the charge +q at the origin is:
E1 = k*q/(x^2)
Similarly, the electric field at any point on the +x axis due to the charge -2q at x=6.30m is:
E2 = k*(-2q)/(x-6.30)^2
Step 2: Set up an equation to find the finite value of x where the electric field is zero.
Since we want to find the value of x where the electric field is zero (E=0), we equate E1 and E2:
k*q/(x^2) = k*(-2q)/(x-6.30)^2
Step 3: Solve the equation to find the finite value of x.
We can cancel out k and q from both sides of the equation:
1/(x^2) = -2/(x-6.30)^2
Cross-multiply the equation:
(x-6.30)^2 = -2x^2
Expand the equation:
x^2 - 12.6x + 39.69 = -2x^2
Bring all terms to one side:
3x^2 - 12.6x + 39.69 = 0
Now you can solve this quadratic equation to find the finite values of x where the electric field is zero.
(b) For what finite values of x is the electric potential zero?
Step 1: Determine the expression for the electric potential at any point on the +x axis due to the charges.
The electric potential at a point due to a point charge q is given by the equation:
V = k*q/r
where V is the electric potential, q is the charge, r is the distance from the charge, and k is the electrostatic constant.
In this case, the electric potential at any point on the +x axis due to the charge +q at the origin is:
V1 = k*q/x
Similarly, the electric potential at any point on the +x axis due to the charge -2q at x=6.30m is:
V2 = k*(-2q)/(x-6.30)
Step 2: Set up an equation to find the finite values of x where the electric potential is zero.
Since we want to find the values of x where the electric potential is zero (V=0), we equate V1 and V2:
k*q/x = k*(-2q)/(x-6.30)
Step 3: Solve the equation to find the finite values of x.
We can cancel out k and q from both sides of the equation:
1/x = -2/(x-6.30)
Cross-multiply the equation:
(x-6.30) = -2x
Expand the equation:
x - 6.30 = -2x
Bring all terms to one side:
3x = 6.30
Now you can solve this equation to find the finite value of x where the electric potential is zero.
Please note that for part (b), the electric potential can only be zero at one point since it is a scalar quantity.
To find the answers to these questions, we need to use the principles and formulas related to electric fields and electric potentials. Here's a step-by-step guide on how to approach these problems:
(a) For what finite value of x is the electric field zero?
Step 1: Understand the problem
We have two charges, +q at the origin (x = 0) and -2q at x = 6.30 m on the +x axis. We need to find the value of x where the electric field is zero.
Step 2: Use Coulomb's Law
The electric field created by a point charge is given by Coulomb's Law:
E = k * (q / r^2)
Where E is the electric field, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
Step 3: Analyze the charges
For the electric field to be zero, the electric fields created by both charges at a particular point should cancel each other out.
The electric field produced by the +q charge at any point on the +x axis is always in the negative x direction. Similarly, the electric field produced by the -2q charge is also in the negative x direction.
Step 4: Set up the equation
Since the electric fields produced by both charges are in the same direction, they will add up algebraically. So, the equation would be:
E_resultant = E_+q + E_-2q = 0
Step 5: Solve for x
Substituting the electric field equation into the above equation, we get:
k * (q / x^2) - k * (2q / (6.30 - x)^2) = 0
Now, we can solve this equation for x.
(b) For what finite values of x is the electric potential zero?
Step 1: Understand the problem
We need to find the values of x where the electric potential is zero. The electric potential at a point due to a single charge is given by:
V = k * (q / r)
Where V is the electric potential, k is the electrostatic constant, q is the charge, and r is the distance from the charge.
Step 2: Electric potential due to +q charge
The electric potential produced by the +q charge at any point on the +x axis can be calculated using the above equation.
Step 3: Electric potential due to -2q charge
Similarly, we need to calculate the electric potential produced by the -2q charge at any point on the +x axis.
Step 4: Setup the equation
For the electric potential to be zero, the electric potentials produced by both charges should cancel each other out at a particular point.
Step 5: Solve for x
Equate the electric potentials calculated for each charge at a particular point to zero and solve for x.
Finally, by following these steps, you should be able to find the finite value(s) of x where the electric field and electric potential are zero.