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A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ?
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A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another...
Problem:
Find the point along the x-axis between two charges where the electric potential resulting from both charges is minimum.
Solution:
Step 1: Understanding the problem
We have two charges, q1 = 0.829 nC and q2 = 0.275 nC, placed at positions r1 = 0 and r2 = 11.9 cm on the x-axis, respectively. We need to find the point between these two charges where the electric potential is minimum.
Step 2: Electric potential due to a point charge
The electric potential (V) at a point in space due to a point charge (q) is given by the formula:
V = k * q / r
where k is the Coulomb constant (k = 9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.
Step 3: Electric potential due to two point charges
When we have more than one charge, the total electric potential at a point is the sum of the electric potentials due to each individual charge. So, the electric potential (V) at a point between two charges (q1 and q2) is given by the formula:
V = k * q1 / r1 + k * q2 / r2
Step 4: Finding the point of minimum electric potential
To find the point along the x-axis where the electric potential is minimum, we need to differentiate the electric potential equation with respect to the position (x) and set it equal to zero. Then, we solve for x.
Let's calculate the electric potential at a point x on the x-axis between the two charges:
V(x) = k * q1 / (x - 0) + k * q2 / (11.9 - x)
Differentiate V(x) with respect to x:
dV/dx = -k * q1 / (x - 0)^2 + k * q2 / (11.9 - x)^2
Set dV/dx equal to zero and solve for x:
-k * q1 / (x - 0)^2 + k * q2 / (11.9 - x)^2 = 0
Simplifying the equation, we get:
q1 / (x - 0)^2 = q2 / (11.9 - x)^2
Cross-multiplying:
q1 * (11.9 - x)^2 = q2 * (x - 0)^2
Expanding and simplifying:
(q1 * (11.9 - x))^2 = (q2 * x)^2
Taking the square root of both sides:
q1 * (11.9 - x) = q2 * x
Simplifying further:
11.9 * q1 - q1 * x = q2 * x
Rearranging the equation:
(11.9 * q1) / (q1 + q2) = x
Substituting the given values:
x = (11.9 * 0.829 nC) / (0.829 nC + 0.275 nC)
x = (11.9 * 0.829) / 1.104
x ≈
Find the point along the x-axis between two charges where the electric potential resulting from both charges is minimum.
Solution:
Step 1: Understanding the problem
We have two charges, q1 = 0.829 nC and q2 = 0.275 nC, placed at positions r1 = 0 and r2 = 11.9 cm on the x-axis, respectively. We need to find the point between these two charges where the electric potential is minimum.
Step 2: Electric potential due to a point charge
The electric potential (V) at a point in space due to a point charge (q) is given by the formula:
V = k * q / r
where k is the Coulomb constant (k = 9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.
Step 3: Electric potential due to two point charges
When we have more than one charge, the total electric potential at a point is the sum of the electric potentials due to each individual charge. So, the electric potential (V) at a point between two charges (q1 and q2) is given by the formula:
V = k * q1 / r1 + k * q2 / r2
Step 4: Finding the point of minimum electric potential
To find the point along the x-axis where the electric potential is minimum, we need to differentiate the electric potential equation with respect to the position (x) and set it equal to zero. Then, we solve for x.
Let's calculate the electric potential at a point x on the x-axis between the two charges:
V(x) = k * q1 / (x - 0) + k * q2 / (11.9 - x)
Differentiate V(x) with respect to x:
dV/dx = -k * q1 / (x - 0)^2 + k * q2 / (11.9 - x)^2
Set dV/dx equal to zero and solve for x:
-k * q1 / (x - 0)^2 + k * q2 / (11.9 - x)^2 = 0
Simplifying the equation, we get:
q1 / (x - 0)^2 = q2 / (11.9 - x)^2
Cross-multiplying:
q1 * (11.9 - x)^2 = q2 * (x - 0)^2
Expanding and simplifying:
(q1 * (11.9 - x))^2 = (q2 * x)^2
Taking the square root of both sides:
q1 * (11.9 - x) = q2 * x
Simplifying further:
11.9 * q1 - q1 * x = q2 * x
Rearranging the equation:
(11.9 * q1) / (q1 + q2) = x
Substituting the given values:
x = (11.9 * 0.829 nC) / (0.829 nC + 0.275 nC)
x = (11.9 * 0.829) / 1.104
x ≈
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A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ?
Question Description
A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ?.
A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charge of q_1 = 0.829 nC is placed at r_1 = 0 on the x-axis. Another charge of q_2 = 0.275 nC is placed at r_2 = 11.9 cm on the x-axis. At which point along the x-axis between the two charges does the electric potential resulting from both of the charges have a minimum ?.
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