Two charges are placed on the x axis. One of the charges (q1 = +8.71C) is at x1 = +3.00 cm and the other (q2 = -28.6C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.
Please help!
To find the net electric field at a particular point, we can use the principle of superposition. Supposing we have multiple point charges, the net electric field at a specific location is the vector sum of the individual electric fields created by each charge.
When calculating the electric field due to a point charge, we can use the following equation:
E = k * |q| / r^2
Where:
E is the electric field
k is Coulomb's constant (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
Let's calculate the net electric field at the given points:
(a) x = 0 cm:
We have two charges q1 = +8.71 C at x1 = +3.00 cm and q2 = -28.6 C at x2 = +9.00 cm.
To find the electric field at x = 0 cm, we need to calculate the individual electric fields due to q1 and q2, and then sum them up.
For q1:
|q1| = 8.71 C
r1 = 3.00 cm = 0.03 m
E1 = k * |q1| / r1^2 = (8.99 x 10^9 Nm^2/C^2) * 8.71 C / (0.03 m)^2
For q2:
|q2| = 28.6 C
r2 = 9.00 cm = 0.09 m
E2 = k * |q2| / r2^2 = (8.99 x 10^9 Nm^2/C^2) * 28.6 C / (0.09 m)^2
Now, we can sum up the electric fields due to q1 and q2 to find the net electric field at x = 0 cm:
Net electric field E = E1 + E2
(b) x = +6.00 cm:
To find the net electric field at x = +6.00 cm, we follow the same process as in (a) but use the new point.
For q1:
|q1| = 8.71 C
r1 = x - x1 = 0.06 m - 0.03 m = 0.03 m (distance between x and x1)
E1 = k * |q1| / r1^2 = (8.99 x 10^9 Nm^2/C^2) * 8.71 C / (0.03 m)^2
For q2:
|q2| = 28.6 C
r2 = x - x2 = 0.06 m - 0.09 m = -0.03 m (distance between x and x2)
E2 = k * |q2| / r2^2 = (8.99 x 10^9 Nm^2/C^2) * 28.6 C / (-0.03 m)^2
Again, sum up E1 and E2 to find the net electric field at x = +6.00 cm.
Remember, in both cases, the direction of the net electric field will be indicated by a plus or minus sign.