What is the sign and magnitude of a point charge that produces a potential of -2.00 V at a distance of 3.00 mm?
To find the sign and magnitude of the point charge, we can use the formula for the potential due to a point charge:
V = k * (|q| / r),
where
V is the potential,
k is the electrostatic constant (k = 9.0 x 10^9 N·m^2/C^2),
|q| is the magnitude of the point charge,
and r is the distance from the charge.
Given that the potential (V) is -2.00 V and the distance (r) is 3.00 mm (which is equivalent to 0.00300 m), we can rearrange the formula to solve for |q|:
|q| = (V * r) / k.
Substituting the given values into the formula:
|q| = (-2.00 V * 0.00300 m) / (9.0 x 10^9 N·m^2/C^2).
Calculating:
|q| = -6.00 x 10^-3 C.
The magnitude of the charge, |q|, is 6.00 x 10^-3 C.
Since the potential is negative, the sign of the point charge is negative, indicating that it is a negative charge.
To determine the sign and magnitude of a point charge that produces a potential of -2.00 V at a distance of 3.00 mm, we can use the equation for electric potential:
V = k * q / r
Where:
V is the electric potential,
k is the electrostatic constant (8.99 × 10^9 N·m^2/C^2),
q is the magnitude of the charge,
and r is the distance from the charge.
Given that V = -2.00 V and r = 3.00 mm = 0.00300 m, we can rearrange the equation to solve for q:
q = V * r / k
Plugging in the values:
q = (-2.00 V) * (0.00300 m) / (8.99 × 10^9 N·m^2/C^2)
By calculating this expression, we can find the magnitude of the charge. Additionally, since the potential is negative, we know that the charge must be opposite in sign to the resulting value of q.