! i have a physics exam tomorrow and i just can't wrap my head around this one!

three charged particles lie on the same straight line. their charges are equal:

qa=qb=qc=0.5 microcoulombs

charged particle b is in between particles a and c, in the middle. what is the intensity of the electrostatic force acting on particle a? the distance between a and c is 20 meters. particles b and c are fixed in place.

formula for the electrostatic force:

the answer is 28 micronewtons, but i don't understand how they got that solution?

Part A - The apparent displacement of the mark is 4 cm.

The refractive index of a material, n = real depth, D/apparent depth, L.

So, n = D/L

Since the refractive index of the triangular glass prism is 1.5, n = 1.5. Also, the triangular glass prism is 12 cm thick and placed on a mark on a piece of paper resting on a horizontal bench. So, the real depth of the mark on the piece of paper through the triangular prism is D = 12 cm and its apparent depth is L.

Since n = D/L,

making L, subject of the formula, we have

L = D/n

Substituting the values of the variables into the equation, we have

L = D/n

L = 12 cm/1.5

L = 8 cm

so, the apparent depth of the mark is 8 cm.

So, the apparent displacement of the mark is d = D - L = 12 cm - 8 cm = 4 cm

Thus the apparent displacement of the mark is 4 cm.

Part B - The refractive index of water, n = 1.33.

The refractive index of a material, n = real depth, D/apparent depth, L.

So, n = D/L

Since the depth of water in the jar is 24 cm, the real depth, D = 24 cm.

Also, the bottom of the jar appears to be raised by 6 cm. So, the apparent depth, L = 24 cm - 6 cm = 18 cm

Since n = D/L, the refractive index of water n = D/L

Substituting the values of the variables into the equation, we have

n = D/L

= 24 cm/18 cm

= 1.33

So, the refractive index of water, n = 1.33.

Learn more about refractive index here: