A 5 C charged is placed on the 0 cm mark of a meter stick and a - 4 C charged is placed on the 50 cm mark what is the electric field at the 30 cm mark
To calculate the electric field at the 30 cm mark on the meter stick due to the charges at the 0 cm and 50 cm marks, you need to use the principle of superposition. The electric field is a vector quantity, meaning it has both magnitude and direction, and it is determined by the charges present in the surrounding space.
Here are the steps to calculate the electric field at the 30 cm mark:
1. Determine the distance between the 5 C charge at the 0 cm mark and the 30 cm mark. In this case, it is 30 cm (or 0.30 m).
2. Calculate the electric field due to the 5 C charge. Use Coulomb's Law, which states that the electric field created by a point charge is given by the equation:
E = k * Q / r^2
where E is the electric field, k is the electrostatic constant (k = 9.0 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance between the charges.
Plugging in the values, we get:
E1 = (9.0 x 10^9 Nm^2/C^2) * (5 C) / (0.30 m)^2
E1 = 300 N/C
So, the electric field due to the 5 C charge at the 0 cm mark is 300 N/C at the 30 cm mark.
3. Determine the distance between the -4 C charge at the 50 cm mark and the 30 cm mark. In this case, it is 20 cm (or 0.20 m).
4. Calculate the electric field due to the -4 C charge. Use Coulomb's Law again:
E2 = (9.0 x 10^9 Nm^2/C^2) * (-4 C) / (0.20 m)^2
E2 = -900 N/C
Note the negative sign indicates the direction of the electric field due to the negative charge.
5. Add the electric fields from both charges together to get the total electric field at the 30 cm mark:
E_total = E1 + E2
E_total = 300 N/C + (-900 N/C)
E_total = -600 N/C
The negative sign indicates that the electric field points in the opposite direction to the positive direction of the meter stick.
Therefore, the electric field at the 30 cm mark on the meter stick due to the charges at the 0 cm and 50 cm marks is -600 N/C.