three point charges are located at the corners of a right triangle where q1=q3= 5uC,Q2= -2uC and r =10cm.(angle q1 q2q3=90 and q1q3=r)
consider 3points charge place at the corners of aright triangle as shown below q1,q2,q3,find the resultant force on q3,a=o.1m,q1=q2=5uc,q3=-2uc
Student
To find the electric field at the remaining corner point of the right triangle, we can use the principle of superposition.
Step 1: Calculate the electric field due to each point charge separately.
The electric field due to a point charge can be calculated using the formula:
E = k * (q / r^2),
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance between the point charge and the point where you want to calculate the electric field.
For q1:
q1 = 5 μC,
r1 = 10 cm = 0.1 m.
E1 = k * (q1 / r1^2)
For q3:
q3 = 5 μC,
r3 = 10 cm = 0.1 m.
E3 = k * (q3 / r3^2)
For q2:
q2 = -2 μC,
r2 = ? (Since the problem does not provide the distance for q2, we cannot calculate its electric field. However, we will be able to calculate the total electric field at the remaining corner point using the electric fields due to q1 and q3.)
Step 2: Calculate the total electric field at the remaining corner point.
Since the electric field due to each charge is a vector quantity, we need to consider both the magnitude and direction of the electric fields.
E_total = E1 + E3
The direction of the electric field due to q1 will be towards q1 (away from the remaining corner point), and the direction of the electric field due to q3 will also be towards q3 (towards the remaining corner point).
To find the direction of the total electric field, we need to determine whether the magnitudes of E1 and E3 are equal or not.
Step 3: Determine the direction of the total electric field.
Since q1 and q3 have the same magnitude (both 5 μC), their electric fields will have the same magnitude as well.
Therefore, the magnitudes of E1 and E3 will add up, and the direction will be determined by the algebraic sum of their magnitudes.
If E1 = E3 (equal magnitudes), the total electric field will be zero (since the magnitudes of the field vectors cancel each other out).
If E1 ≠ E3 (unequal magnitudes), the total electric field will have a non-zero magnitude and a direction determined by the difference in the magnitudes of E1 and E3.
To determine the direction, compare the magnitudes of E1 and E3 to see which one is larger.
To calculate the electric potential energy between the three point charges, we can use the formula:
PE = k * (|q1*q2|/r12 + |q1*q3|/r13 + |q2*q3|/r23)
Where:
- PE represents the electric potential energy
- k is the electrostatic constant, which is approximately 9 x 10^9 Nm^2/C^2
- q1, q2, and q3 are the magnitudes of the charges
- r12, r13, and r23 are the distances between the charges
Given:
- q1 = q3 = 5 uC (microCoulombs), which is equivalent to 5 x 10^-6 C
- q2 = -2 uC, which is equivalent to -2 x 10^-6 C
- r = 10 cm, which is equivalent to 0.1 m
First, let's calculate the distance between the charges:
Since the angle between q1 and q3 is 90 degrees and q1q3 = r, we have a right triangle.
Using the Pythagorean theorem, we can find the distance between q1 and q3 (r13).
r13 = sqrt(r^2 + r^2) = sqrt(0.1^2 + 0.1^2) = sqrt(0.02) = 0.1414 m
Now, we can calculate the electric potential energy:
PE = k * (|q1*q2|/r12 + |q1*q3|/r13 + |q2*q3|/r23)
= (9 x 10^9 Nm^2/C^2) * (|5x10^-6 C * -2x10^-6 C| / (0.1 m) + |5x10^-6 C * 5x10^-6 C| / (0.1414 m) + |-2x10^-6 C * 5x10^-6 C| / (0.1414 m))
Simplifying,
PE = (9 x 10^9 Nm^2/C^2) * (10^-5 / 0.1 + 25 x 10^-12 / 0.1414 - 10^-11 / 0.1414)
= (9 x 10^9) * (10^-4 + 1.7725 x 10^-10 - 7.071 x 10^-10)
= 9 x 10^9 * (-5.831 x 10^-10)
≈ -5.25 Joules
Therefore, the electric potential energy between the three point charges is approximately -5.25 Joules.