Two point charges, 3.4 μC and -2.6 μC, are placed 1.6 cm apart on the x-axis. At what points along the x-axis is (a) the electric field zero (b) the potential zero?
two point charges 8*10c and 4*10c are 10cm apart.calculate the word done in bringing them 4cm closer.
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To find the points along the x-axis where the electric field is zero, we can use the principle that the electric field due to a point charge is given by the equation:
E = k * q / r^2,
where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the point charge.
(a) The electric field will be zero at some point on the x-axis if the electric fields due to the two charges cancel each other out. So we need to find the point where the electric field due to the charge of 3.4 μC is equal in magnitude but opposite in direction to the electric field due to the charge of -2.6 μC.
Let's call the distance from the point charge of 3.4 μC to the point on the x-axis x, and the distance from the point charge of -2.6 μC to the same point on the x-axis as (1.6 cm - x). Using the electric field equation for each charge, we can set them equal to each other:
k * 3.4 μC / x^2 = k * (-2.6 μC) / (1.6 cm - x)^2.
Simplifying this equation, we get:
3.4 / x^2 = -2.6 / (1.6 - x)^2.
Cross multiplying and expanding, we have:
3.4 * (1.6 - x)^2 = -2.6 * x^2.
Expanding further, we get:
5.44 - 6.8x + 2.72x^2 = -2.6x^2.
Combining like terms, we have:
2.72x^2 + 6.8x - 5.44 = 0.
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a),
where a = 2.72, b = 6.8, and c = -5.44.
Solving, we get two possible solutions for x: x ≈ -1.19 cm or x ≈ 0.62 cm.
Since we are considering the x-axis, the negative value (-1.19 cm) is not physically meaningful. Therefore, the electric field is zero at approximately x = 0.62 cm on the x-axis.
(b) The potential due to a point charge is given by the equation:
V = k * q / r,
where V is the potential, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the point charge.
The potential is zero at a point if the potentials due to the two charges cancel each other out. We need to find the point where the potential due to the charge of 3.4 μC is equal in magnitude but opposite in sign to the potential due to the charge of -2.6 μC.
Using the potential equation for each charge, we can set them equal to each other:
k * 3.4 μC / x = k * (-2.6 μC) / (1.6 cm - x).
Simplifying this equation, we get:
3.4 / x = -2.6 / (1.6 - x).
Cross multiplying and simplifying, we have:
3.4 * (1.6 - x) = -2.6 * x.
Expanding and simplifying further, we get:
5.44 - 3.4x = -2.6x.
Combining like terms, we have:
0.8x = 5.44.
Dividing both sides by 0.8, we find:
x = 6.8.
Therefore, the potential is zero when x = 6.8 cm on the x-axis.
To determine the points along the x-axis where the electric field and potential are zero, we need to consider the principle of superposition, which states that the total electric field and potential at any point due to multiple charges is the vector sum of the electric fields and potentials produced by each individual charge.
Let's tackle each part separately:
(a) Electric Field Zero:
To find the points along the x-axis where the electric field is zero, we need to find the locations where the vector sum of the electric fields produced by each point charge is zero. Mathematically, this can be represented as:
E_total = E1 + E2
where E1 and E2 are the electric fields produced by charges 3.4 μC and -2.6 μC, respectively.
The electric field at a point 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, and r is the distance between the charge and the point of interest.
To find the points where the electric field is zero, we need to set up the equation:
E1 + E2 = 0
k * (q1 / r1^2) + k * (q2 / r2^2) = 0
Substituting the given values, we get:
(8.99 x 10^9 Nm^2/C^2) * (3.4 x 10^-6 C / r1^2) + (8.99 x 10^9 Nm^2/C^2) * (-2.6 x 10^-6 C / r2^2) = 0
Simplifying, we have:
3.4 / r1^2 - 2.6 / r2^2 = 0
By solving this equation, you can find the values of r1 and r2, which represent the distances from the respective charges to the points where the electric field is zero along the x-axis.
(b) Potential Zero:
To find the points along the x-axis where the potential is zero, we need to find the locations where the sum of the potentials produced by each point charge is zero. The potential at a point due to a point charge can be calculated using the equation:
V = k * (q / r)
where V is the potential, k is the Coulomb's constant, q is the charge, and r is the distance between the charge and the point of interest.
Using the principle of superposition, we can set up the equation:
V_total = V1 + V2 = 0
k * (q1 / r1) + k * (q2 / r2) = 0
Substituting the given values, we have:
(8.99 x 10^9 Nm^2/C^2) * (3.4 x 10^-6 C / r1) + (8.99 x 10^9 Nm^2/C^2) * (-2.6 x 10^-6 C / r2) = 0
Simplifying, we get:
3.4 / r1 - 2.6 / r2 = 0
By solving this equation, you can find the values of r1 and r2, which represent the distances from the respective charges to the points where the potential is zero along the x-axis.