What would be the electric field (magnitude and direction) of 1.50 cm to the right of a charge of -6.5 × 10-6 C?
plz answer as fast as possible
To calculate the electric field at a point, you can use the formula for electric field strength:
Electric field = (k * q) / r^2
Where:
- k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- q is the charge
- r is the distance from the charge to the point where you want to calculate the electric field.
In this case, the charge is -6.5 × 10^-6 C, and the distance from the charge to the point of interest is 1.50 cm, which we need to convert to meters (1 cm = 0.01 m).
So, the first step is to calculate the electric field:
Electric field = (9 x 10^9 Nm^2/C^2 * (-6.5 × 10^-6 C)) / (0.01 m)^2
Calculating this equation will give you the magnitude of the electric field at the point to the right of the charge.
Now, since the charge is negative, the electric field will point towards it in the opposite direction. In this case, to the left.
Therefore, the magnitude of the electric field is the result of the above calculation, and the direction of the electric field is to the left.
d = 0.0150 meter
E = k q / d^2