Question

two point charges two point charges are separated by 25 cm in the figure find The Net electric field these charges produced at point a and point b ​

Answer 1

The student's question involves calculating the zero electric field point between two point charges, determining the electric field at the midpoint between two charges, and sketching electric field line maps for pairs of charges both alike and opposite.

Step-by-step explanation:

The student is dealing with concepts related to electric fields created by point charges. The task involves calculating net electric fields at certain points relative to the charges and understanding how electric field lines map out around different configurations of charges.

Zero Electric Field Point

To find the point where the electric field is zero between two charges (25.0 µC and 45.0 µC placed 0.500 m apart), you must set the magnitudes of the electric fields due to each charge equal to each other and solve for the distance from one of the charges to the zero field point. This usually results in one of the charges needing to be much closer to the zero-field point than the other due to its smaller magnitude.

Electric Field Halfway

At the halfway point between two charges, if they are of opposite signs and equal in magnitude, the net electric field would be zero. However, if the charges are not equal, you can determine the electric field at the halfway point by finding the vector sum of the electric fields due to each charge at that point. The direction of the vector will depend on the sign of each charge.

Electric Field Lines Map

When drawing the electric field lines for two like charges, the lines repel each other and the field is strongest between the charges. For opposite charges, the lines are directed from the positive charge towards the negative charge, and they illustrate the direction in which a positive test charge would move. A map for two charges of +20 µC and +20 µC will show lines emanating outwards, whereas for +20 µC and -30 µC charges, the lines will begin at the positive charge and end at the negative charge.

Answer 2

The image is missing and so i have attached it.

Answer:

A) E = 8740 N/C

B) E = -6536 N/C

Step-by-step explanation:

The formula for electric field is;

E = kq/r²

Where;

q is charge

k is a constant with value 8.99 x 10^(9) N•m²/C²

A) Now, to find the net electric field at point A, the formula would now be;

E = (kq1/(r1)²) - (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point A

r2 is distance from charge q2 to point A.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m

r2 = 10 cm = 0.1 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))

E = 8740 N/C

B) similarly, electric field at point B;

E = (kq1/(r1)²) + (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point B

r2 is distance from charge q2 to point B.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 10 cm = 0.1 m

r2 = 25cm + 10 cm = 35 cm = 0.35 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))

E = -6536 N/C