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 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 electrostatic potential:
V = k * Q / r
where:
V is the electrostatic potential (in volts),
k is the Coulomb's constant (k = 8.99 * 10^9 Nm^2/C^2),
Q is the charge, and
r is the distance from the charge.
In this case, we are given:
V = -2.00 V (negative potential),
r = 3.00 mm (distance in meters), and
k = 8.99 * 10^9 Nm^2/C^2.
We can rearrange the equation to solve for Q:
Q = V * r / k
Substituting the given values, we have:
Q = (-2.00 V) * (3.00 * 10^(-3) m) / (8.99 * 10^9 Nm^2/C^2)
Calculating this expression, we get:
Q ≈ -6.68 * 10^(-10) C
The magnitude of the charge is 6.68 * 10^(-10) C. Since the potential is negative, we can conclude that the charge is also negative.
To determine the sign and magnitude of the point charge, we can use the equation for electric potential:
V = k * q / r
where:
V is the potential,
k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2),
q is the charge, and
r is the distance from the charge.
Rearranging the equation to solve for the charge:
q = V * r / k
Substituting the given values:
V = -2.00 V (negative since potential is negative),
r = 3.00 mm (or 3.00 x 10^-3 m),
k = 8.99 x 10^9 N m^2/C^2:
q = (-2.00 V) * (3.00 x 10^-3 m) / (8.99 x 10^9 N m^2/C^2)
Calculating the value:
q = -6.00 x 10^-3 m²/C
Therefore, the sign of the point charge is negative, and the magnitude of the charge is 6.00 x 10^-3 C.