Three charges q1= 2uc(0,0)cm q2=4uc(0,10)cm and q3=_3uc(15,0)cm acts on one another, find the magnitude of the total electric force and the angle that the total electric force makes with positive x_axis.
two forces are in effect
Fnet=kq1q2/.1^2 + kq1q3/.15^2
the first is directed perpendicular to x axis. The other, has a component perpendicular to the x axis of kq1q3/d^2*cosTheta where d is sqrt(.1^2+.15^2) and cosTheta=.1/(d)
total force on x axis: kq1q2/.1^2 + kq1q3/(d^2) * .1/d
To find the magnitude of the total electric force and the angle it makes with the positive x-axis, we can use the concept of vector addition. The electric force between two charges is given by Coulomb's law:
F = k * |q1 * q2| / r^2
Where F is the magnitude of the electric force, k is the electrostatic constant (9 × 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Let's calculate the electric forces between each pair of charges:
For q1 and q2, the distance between them is given by:
r12 = √((x2 - x1)^2 + (y2 - y1)^2)
= √(0^2 + (10 - 0)^2)
= √(100)
= 10 cm
The electric force between q1 and q2 is:
F12 = k * |q1 * q2| / r12^2
= (9 × 10^9 Nm^2/C^2) * |2 × 10^-6 C * 4 × 10^-6 C| / (10 × 10^-2 m)^2
= (9 × 10^9 Nm^2/C^2) * (8 × 10^-12 C^2) / 10^-2 m^2
= 72 × 10^-3 N
For q1 and q3, the distance between them is given by:
r13 = √((x3 - x1)^2 + (y3 - y1)^2)
= √((15 - 0)^2 + 0^2)
= √(225)
= 15 cm
The electric force between q1 and q3 is:
F13 = k * |q1 * q3| / r13^2
= (9 × 10^9 Nm^2/C^2) * |2 × 10^-6 C * -3 × 10^-6 C| / (15 × 10^-2 m)^2
= (9 × 10^9 Nm^2/C^2) * (-6 × 10^-12 C^2) / (15 × 10^-2 m)^2
= -24 × 10^-3 N
For q2 and q3, the distance between them is given by:
r23 = √((x3 - x2)^2 + (y3 - y2)^2)
= √((15 - 0)^2 + (0 - 10)^2)
= √(225 + 100)
= √(325)
≈ 18.02 cm
The electric force between q2 and q3 is:
F23 = k * |q2 * q3| / r23^2
= (9 × 10^9 Nm^2/C^2) * |4 × 10^-6 C * -3 × 10^-6 C| / (18.02 × 10^-2 m)^2
= (9 × 10^9 Nm^2/C^2) * (-12 × 10^-12 C^2) / (18.02 × 10^-2 m)^2
= -27 × 10^-3 N
Now let's find the components of the total electric force in the x and y directions by summing up the electric force vectors:
Fx = F12x + F13x + F23x
= 0 + 0 + 27 × 10^-3 N
= 27 × 10^-3 N
Fy = F12y + F13y + F23y
= 72 × 10^-3 N + (-24 × 10^-3 N) + 0
= 48 × 10^-3 N
To calculate the magnitude of the total electric force (F) and the angle (θ) it makes with the positive x-axis, we can use the Pythagorean theorem and trigonometry. The magnitude is given by:
F = √(Fx^2 + Fy^2)
= √((27 × 10^-3 N)^2 + (48 × 10^-3 N)^2)
≈ 55.80 × 10^-3 N
To calculate the angle θ, we can use the formula:
θ = tan^(-1)(Fy / Fx)
= tan^(-1)((48 × 10^-3 N) / (27 × 10^-3 N))
= tan^(-1)(1.78)
≈ 61.28°
Therefore, the magnitude of the total electric force is approximately 55.80 × 10^-3 N, and the angle it makes with the positive x-axis is approximately 61.28°.