Two point charges lie along the y axis. A charge of q1 = -12.0 µC is at y = 6.0 m, and a charge of q2 = -5.0 µC is at y = -3.0 m. Locate the point (other than infinity) at which the total electric field is zero.
P=7
To find the point where the total electric field is zero, we can use the principle of superposition.
The electric field at a point due to a charge is given by Coulomb's Law:
E = k * q / r^2,
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance between the point and the charge.
Since both charges are negative (q1 = -12.0 µC and q2 = -5.0 µC), the electric fields they produce will be directed towards them.
Let's consider an arbitrary point on the y-axis, denoted as y. The electric field at this point due to q1 is directed towards the negative y-direction and its magnitude is given by:
E1 = k * |q1| / (y - 6.0)^2.
Similarly, the electric field at this point due to q2 is directed towards the negative y-direction and its magnitude is given by:
E2 = k * |q2| / (y - (-3.0))^2.
To find the point where the total electric field is zero, we need the sum of these two electric fields to be equal to zero:
E_total = E1 + E2 = 0.
Substituting the expressions for E1 and E2:
k * |q1| / (y - 6.0)^2 + k * |q2| / (y + 3.0)^2 = 0.
Now, let's solve this equation algebraically to find the value of y where the total electric field is zero.
To find the point at which the total electric field is zero, we need to calculate the electric field created by each charge individually and then determine the point where the vector sum of the electric fields is zero.
The electric field created by a point charge is given by the equation:
E = k * (q / r^2) * u
where,
E is the electric field,
k is the Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2),
q is the magnitude of the charge,
r is the distance from the charge to the point where the electric field is being calculated,
u is the unit vector pointing from the charge to the point.
Let's calculate the electric fields produced by each charge at a general point on the y-axis, located at (0, y).
For the charge q1 at y = 6.0 m:
Distance from q1 to the point = (0 - 6.0) m = -6.0 m
Electric field created by q1 at (0, y) is given by:
E1 = k * (q1 / r1^2) * u1
= (8.99 x 10^9 Nm^2/C^2) * (-12.0 x 10^-6 C) / (-6.0 m)^2 * u1
= -2.498 x 10^6 N/C * u1
For the charge q2 at y = -3.0 m:
Distance from q2 to the point = (0 - (-3.0)) m = 3.0 m
Electric field created by q2 at (0, y) is given by:
E2 = k * (q2 / r2^2) * u2
= (8.99 x 10^9 Nm^2/C^2) * (-5.0 x 10^-6 C) / (3.0 m)^2 * u2
= -5.997 x 10^6 N/C * u2
To find the point where the total electric field is zero, we need to determine the values of y for which E1 + E2 = 0.
Since the electric fields E1 and E2 are vectors, they can be added using vector addition.
E1 + E2 = -2.498 x 10^6 N/C * u1 + (-5.997 x 10^6 N/C * u2)
To find the point where E1 + E2 = 0, we need to solve the vector equation:
-2.498 x 10^6 N/C * u1 + (-5.997 x 10^6 N/C * u2) = 0
By solving this equation, we can determine the value of y at the point where the total electric field is zero.
some point (0,y) on y axis between
y = -3 and y = +6
field up = k(-12)/(6-y)^2
field down = k(-5)/(y-3)^2
so
12(y-3)^2 = 5(6-y)^2