Two charges 10uC (point A) and -10uC (point B) are placed 10cm from each other. Find the electric field at a point P perpendicular bisector of AB at a distance 12cm from the middle.
Can someone get me started on this one? I'd appreciate any help. Thanks
Sure. The total electric field is a vector combination of the two. Now if you sketch the directions, you will note the vertical component of the + charge is equal and opposite to the vertical component of the - charge, so the net E is zero in the vertical. Now in the horizontal direction, they are equal, and in the same direction. So take the E of one, multiply it by the sin/cos function to get the horizontal component, then double it. Symettry is a wonderful thing to see.
two charges +20 microculomb and -10 microculomb are 6cm.apart find electric potanial at a pont distance 4 cm.on the right bisector of the line joining the charge.
To find the electric field at point P due to the two charges, we can use Coulomb's Law and the principle of superposition.
1. Determine the distance of point P from each charge: Point P is the perpendicular bisector of AB, so it is equidistant from both charges A and B. Therefore, the distance from P to A is 5 cm (half of the distance between A and B, which is 10 cm) and the distance from P to B is also 5 cm.
2. Calculate the electric field at P due to each charge individually:
a. Electric field due to charge A at point P:
The electric field at point P due to charge A can be calculated using Coulomb's Law:
E1 = k * (q1 / r1^2), where k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2),
q1 is the magnitude of charge A (10 uC = 10 x 10^-6 C), and r1 is the distance from A to P (5 cm = 0.05 m).
b. Electric field due to charge B at point P:
Similar to above, the electric field at point P due to charge B can be calculated as:
E2 = k * (q2 / r2^2), where q2 is the magnitude of charge B (-10 uC = -10 x 10^-6 C), and
r2 is the distance from B to P (also 5 cm = 0.05 m).
3. Apply the principle of superposition:
The total electric field at P is the vector sum of the electric fields due to charges A and B. Since the charges are of equal magnitude and opposite sign, the direction of the resultant electric field will be from B towards A.
E_total = E1 + E2
4. Calculate the magnitude and direction of the total electric field at P:
Substitute the values calculated in step 2 into the equation in step 3 to find the net electric field at P.
Please note that uC refers to microcoulombs and 1 uC is equal to 10^-6 C.
To find the electric field at point P, we can use the principle of superposition.
The electric field due to a point charge can be calculated using Coulomb's law:
E = k * q / r^2
Where E is the electric field, k is the electrostatic constant (8.99 * 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to calculate the electric field.
In this case, we have two charges: 10uC at point A and -10uC at point B. The electric field at point P will be the vector sum of the electric fields due to these charges individually.
Step 1: Calculating the electric field due to the charge at point A:
The distance from point A to point P is 12cm = 0.12m.
Substituting the values into Coulomb's law:
E_A = (k * q_A) / r_A^2
= (8.99 * 10^9 Nm^2/C^2) * (10 * 10^-6 C) / (0.12m)^2
Step 2: Calculating the electric field due to the charge at point B:
The distance from point B to point P is also 12cm = 0.12m.
Since the charge at point B is -10uC, the electric field due to this charge will be in the opposite direction to the electric field at point A. So, we have:
E_B = -(k * q_B) / r_B^2
= -[(8.99 * 10^9 Nm^2/C^2) * (10 * 10^-6 C) / (0.12m)^2]
Step 3: Adding the electric fields:
Since the electric field is a vector, we need to add the magnitudes and consider the direction. At point P, since we have point charges with opposite signs at the same distance, the magnitudes of the electric fields will be the same.
E = E_A + E_B
Keep in mind that the direction of the electric field is from positive charges to negative charges. Therefore, the direction of the electric field at point P will be towards point A.
Now you can substitute the values calculated in Step 1 and Step 2 to find the electric field at point P.