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Three charges –q, Q and –q are placed in a straight line maintaining equal distance from each other. What should be the ratio q/Q so that the net electric potential of the system is zero?
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Three charges –q, Q and –q are placed in a straight line m...
Assuming you are asking about three point charges in space:
Three point charges are placed in space as follows: q1 = +2.0 μC, q2 = -3.0 μC, and q3 = +4.0 μC. The charges are located at the vertices of an equilateral triangle with sides of length 0.50 m.
a) What is the electric potential energy of the system?
b) What is the magnitude and direction of the net force on q1?
c) At what point on the x-axis (other than infinity) is the electric potential zero?
Solution:
a) The electric potential energy of a system of point charges is given by the formula:
U = k * (q1 * q2 / r12 + q1 * q3 / r13 + q2 * q3 / r23)
where k is the Coulomb constant, q1, q2, and q3 are the charges, and r12, r13, and r23 are the distances between them. In this case, the charges are located at the vertices of an equilateral triangle with sides of length 0.50 m. The distances between them are all equal and can be calculated using the Pythagorean theorem:
r12 = r13 = r23 = 0.50 m
Substituting the values into the formula, we get:
U = 9.0 * 10^9 * [(2 * 10^-6 * (-3 * 10^-6)) / 0.50 + (2 * 10^-6 * 4 * 10^-6) / 0.50 + (-3 * 10^-6 * 4 * 10^-6) / 0.50]
U = -3.96 * 10^-5 J
Therefore, the electric potential energy of the system is -3.96 * 10^-5 J.
b) The net force on q1 is the vector sum of the forces due to q2 and q3. The force between two point charges is given by Coulomb's law:
F12 = k * (q1 * q2 / r12^2) * u12
where u12 is the unit vector pointing from q1 to q2. Similarly, the force between q1 and q3 is given by:
F13 = k * (q1 * q3 / r13^2) * u13
where u13 is the unit vector pointing from q1 to q3. The magnitude and direction of these forces can be calculated using the formula:
|F| = k * (|q1| * |q2| / r^2)
where r is the distance between the charges and |q| is the absolute value of the charge. The unit vectors can be calculated using the formula:
u = (r2 - r1) / |r2 - r1|
where r1 and r2 are the position vectors of the charges.
Substituting the values, we get:
F12 = 2.07 * 10^-3 N, u12 = (-i + √3 * j) / 2
F13 = 3.10 * 10^-3 N, u13 = (i + √3 * j) / 2
The net force on q1 is the vector sum of F12 and F13:
Three point charges are placed in space as follows: q1 = +2.0 μC, q2 = -3.0 μC, and q3 = +4.0 μC. The charges are located at the vertices of an equilateral triangle with sides of length 0.50 m.
a) What is the electric potential energy of the system?
b) What is the magnitude and direction of the net force on q1?
c) At what point on the x-axis (other than infinity) is the electric potential zero?
Solution:
a) The electric potential energy of a system of point charges is given by the formula:
U = k * (q1 * q2 / r12 + q1 * q3 / r13 + q2 * q3 / r23)
where k is the Coulomb constant, q1, q2, and q3 are the charges, and r12, r13, and r23 are the distances between them. In this case, the charges are located at the vertices of an equilateral triangle with sides of length 0.50 m. The distances between them are all equal and can be calculated using the Pythagorean theorem:
r12 = r13 = r23 = 0.50 m
Substituting the values into the formula, we get:
U = 9.0 * 10^9 * [(2 * 10^-6 * (-3 * 10^-6)) / 0.50 + (2 * 10^-6 * 4 * 10^-6) / 0.50 + (-3 * 10^-6 * 4 * 10^-6) / 0.50]
U = -3.96 * 10^-5 J
Therefore, the electric potential energy of the system is -3.96 * 10^-5 J.
b) The net force on q1 is the vector sum of the forces due to q2 and q3. The force between two point charges is given by Coulomb's law:
F12 = k * (q1 * q2 / r12^2) * u12
where u12 is the unit vector pointing from q1 to q2. Similarly, the force between q1 and q3 is given by:
F13 = k * (q1 * q3 / r13^2) * u13
where u13 is the unit vector pointing from q1 to q3. The magnitude and direction of these forces can be calculated using the formula:
|F| = k * (|q1| * |q2| / r^2)
where r is the distance between the charges and |q| is the absolute value of the charge. The unit vectors can be calculated using the formula:
u = (r2 - r1) / |r2 - r1|
where r1 and r2 are the position vectors of the charges.
Substituting the values, we get:
F12 = 2.07 * 10^-3 N, u12 = (-i + √3 * j) / 2
F13 = 3.10 * 10^-3 N, u13 = (i + √3 * j) / 2
The net force on q1 is the vector sum of F12 and F13:
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Three charges –q, Q and –q are placed in a straight line m...
Let the distance between any two charges is d. Therefore the net potential energy of the system is 
But the total energy of the system is zero. So, -qQ-qQ+ q2/2
But the total energy of the system is zero. So, -qQ-qQ+ q2/2 =0 that means q=4Q, i.e. q/Q=4.
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