Electric Field due to two point charges r

Electric Field due to two point charges r 1 = [(xf -x 1)2 + (yf -y 1)2]1/2 r 2 = [(x 2 –xf)2 + (y 2 -yf)2]1/2 θ 1= tan-1[(yf -y 1)/(xf -x 1)] θ 2= tan-1[(y 2 -yf)/(x 2 -xf)] note that E 1 is directed away from +q 1 and E 2 is directed towards –q 2 q 1 at (x 1, y 1) + y 1 Note that in determining the q 1 for the positive charge we use (yf – y 1) and (xf – x 1), but since the field points towards the negative charge, we reverse this and use (y 2 – yf) and (x 2 – xf). x r 1 E 2 field point at (xf, yf) E 1 1 r 2 2 - q 2 at (x 2, y 2) This is the slide from Set 1 slide #39.

Example: Electric Field due to two point charges Consider two point charges, a positive charge of 5 n. C located at (-5 cm, +3 cm) and a negative charge of -8 n. C located at (+7 cm, -4 cm). What is the electric field at the location (-2 cm, 6 cm)? First let’s picture the situation (draw a diagram).

Example: Electric Field due to two point charges Consider two point charges, a positive charge of 5 n. C located at (-5 cm, +3 cm) and a negative charge of -8 n. C located at (+7 cm, -4 cm). What is the electric field at the location (-2 cm, -6 cm)? [+5 n. C] q 1 + at (-5 cm, +3 cm) y 1 x r 1 r 2 field point at (-2 cm, -6 cm) 2 1 2 - q 2 [-8 n. C] at (+7 cm, -4 cm)

Example: Electric Field due to two point charges The positive charge causes an electric field to point away from it, and the negative charge causes an electric field to point towards it. To find the total electric field, we need to add the two electric field vectors together – in rectangular form. But first, we’ll determine the field vectors in polar form since that is the way they are usually expressed. y [+5 n. C] q 1 + at (-5 cm, +3 cm) 1 x r 1 E 2 field point at (-2 cm, -6 cm) E 1 1 r 2 2 2 - q 2 [-8 n. C] at (+7 cm, -4 cm)

Example: Electric Field due to two point charges In polar form, the magnitude of E 1 = k* q 1 /r 12 where r 1 = [(x 1 -xf)2 + (y 1 -yf)2]½ = [(-5 cm - -2 cm)2 + (3 cm - -6 cm)2]½ = [(-3 cm)2 + (9 cm)2]½ = [90 cm 2]½ = 9. 49 cm =. 0949 m. E 1 = (9 x 109 Nt*m 2/C 2)*(5 x 10 -9 C)/(. 0949 m)2 = 5, 000 Nt/C. 1 is directed down and right (fourth quadrant): 1 = tan-1{(yf-y 1)/(xf-x 1)} = tan-1{(-6 cm – +3 cm)/(-2 cm - -5 cm)} [+5 n. C] q 1 + y = tan-1{(-9 cm)/(+3 cm)} = -71. 56 o. 1 at (-5 cm, +3 cm) x r 1 E 2 field point 1 at (-2 cm, -6 cm) E 1 r 2 2 2 - q 2 [-8 n. C] at (+7 cm, -4 cm)

Example: Electric Field due to two point charges In polar form, the magnitude of E 2 = k* q 2 /r 22 where r 2 = [(xf-x 2)2 + (yf-y 2)2]½ = [(+7 cm - -2 cm)2 + (-4 cm - -6 cm)2]½ = [(9 cm)2 + (2 cm)2]½ = [85 cm 2]½ = 9. 22 cm =. 0922 m. E 2 = (9 x 109 Nt*m 2/C 2)*(8 x 10 -9 C)/(. 0922 m)2 = 8, 471 Nt/C. 2 is directed up and right (first quadrant): 2 = tan-1{(y 2 -yf)/(x 2 -xf)} = tan-1{(-4 cm – -6 cm)/(+7 cm - -2 cm)} [+5 n. C] q 1 + y = tan-1{(+2 cm)/(+9 cm)} = 12. 53 o. 1 at (-5 cm, +3 cm) x r 1 E 2 field point 1 at (-2 cm, -6 cm) E 1 r 2 2 2 - q 2 [-8 n. C] at (+7 cm, -4 cm)

Example: Electric Field due to two point charges Notice three things: 1. In determining the magnitude in polar form, we used the absolute value of the charge, i. e. , we didn’t include the negative sign of q 2 when calculating the magnitude of E 2. In polar form, magnitudes are always positive (or zero). 2. In determining the value of r, the order of charge point and field point didn’t matter since the difference is squared, i. e. , (xf -x 1)2 = (x 1 -xf)2. 3. In determining the direction for the vectors, we used (field – charge) values if the charge were positive (as is q 1), but we used (charge – field) values if the charge were negative (as is q 2). This is where the sign of the charge comes into play – in determining the direction of the electric field vector.

Example: Electric Field due to two point charges E 1 = 5, 000 Nt/C , -71. 56 o and E 2 = 8, 471 Nt/C , 12. 53 o Notice as a check that the directions of the two vectors are in the quadrants we expected them to be in: E 1 is in the fourth quadrant, E 2 is in the first quadrant. But these are in polar form. We need to add these two electric field vectors in rectangular form. [+5 n. C] q 1 + y 1 at (-5 cm, +3 cm) x r 1 E 2 field point at (-2 cm, -6 cm) E 1 1 r 2 2 2 - q 2 [-8 n. C] at (+7 cm, -4 cm)

Example: Electric Field due to two point charges E 1 = 5, 000 Nt/C , -71. 56 o and E 2 = 8, 471 Nt/C , 12. 53 o To get the total electric field at our chosen field point, we add the electric fields from each of our two charges (in rectangular form): Ex = E 1 cos( 1) + E 2 cos( 2) = 5, 000 Nt/C cos(-71. 56 o) + 8, 471 Nt/C cos(12. 53 o) = 1, 582 Nt/C + 8, 269 Nt/C = 9, 851 Nt/C Ey = E 1 sin( 1) + E 2 sin( 2) = 5, 000 Nt/C sin(-71. 56 o) + 8, 471 Nt/C sin(12. 53 o) = -4, 743 Nt/C + 1, 838 Nt/C = -2, 905 Nt/C

Example: Electric Field due to two point charges Ex = 9, 851 Nt/C Ey = -2, 905 Nt/C Converting back to polar form: E = [Ex 2 + Ey 2]½ = [(9, 851 Nt/C)2 + (-2, 905 Nt/C)2]½ = 10, 270 Nt/C; = tan-1[Ey/Ex] = tan-1 [(-2, 905 Nt/C)/ (9, 851 Nt/C)] = -16. 43 o. Since Ex is positive and Ey is negative, we expect to be in the fourth quadrant.