P118 week 2 ws1 We will now derive an expression for the electric field of a dip

Question

P118 week 2 ws1 We will now derive an expression for the electric field of a dipole for any point on the axis bisecting the dipole (x-axis, see diagram). Please notice that the origin is in the middie between the charges. We will use vector addition of the electric fields due to the negative charge E and due to the positive charge E. Each field is just the field due to a point i charge: E-K Notice that we would get the same result, if the x-axis would point out of the page or in any direction perpendicular to the dipole axis (y-axis here). These points are all on a plane bisecting the dipole. Q26.6 a) Express r' using x and y. Notice that y·s/2 is constant for all points on the x-axis b) Write the magnitudes of the electric fields due to each charge: c) What is the x-component of the electric field at P due to the two charges? two Omne d) What is the y-component of the electric field at P due to the two gharges? 2f el write the electric feld a P sin n vtor notation Since Pis ncfiee fora e) Write the electric field at P using unit vector notation. Since P is not specified, this is the field for any point on x-axis dir -

Explanation / Answer

f)

field is given by

E = 2kq (s/2) /(x^2 +0.5s^2)^1.5

if x>>>>s

E = kqs/x^3

g)

due to point charge

field is kq/x^2

there fore field due to dipole is less than point charge

LESS THAN

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