Two fixed charges, -4MuC and -5. MuC, are separated by a certain distance. If th

Question

Two fixed charges, -4MuC and -5. MuC, are separated by a certain distance. If the charges are separated by 20 cm, calculate the magnitude of the net electric field halfway between the charges. Q1 = -4 x 10^-6 C E = K Q/r^2 Q_2 = -5 x 10^-6 C K = 9 x 10^9 Nm^2/C^2 r = 20cm/2 = 0.1 m E = K Q/r^2 Sigma E_net = E1 + E2 1st. Determine the Electric field and direction of the test charge towards charge towards charge 1 2nd- Determine the Electric field and direction of test charge towards charge 2 3rd-Calculate the net charge by determine the difference or summation of the 2 electric fields.

Explanation / Answer

here,

Electric field at point is given as:
E = K*q/r^2

where,
k = columb's constant
q = charge
r = seperation distance, 20/2 = 10 = 0.1 m

Assuming the mid point to be origin so q1 will on left while q2 on right of origin

For Charge Q1,
E1 = (9*10^9 * -4*10^-6)/(0.1)^2
E1 = -3600 kN/C
- ve sign away from origin

For Charge Q2,
E2 = -(9*10^9 * -5*10^-6)/(0.1)^2
E2 = 4500 kN/C
+ ve sign towards origin

Net Electric field,
En = E1 + E2
En = -3600 + 4500
En = 900 kN/C towards origin or midpoint of charge q1 and q2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.