Question

Please help me out + tell me what I'm doing wrong.
12a: Two charges are held in place as shown. Determine the magnitude and direction of
the electric field at the origin. Use Q = 3 nC, q = −2 nC, D = 4 m, L = 2 m and H = 5 m.
b: Determine the magnitude and direction of the acceleration of a 2 µC charge of mass 10−25
kg which is released at the origin.

Answer 1

For Q:
\[E_{Q}=\frac{8.99*10^{9}Nm^{2}}{C^{2}}*\frac{3nC}{(4m)^{2}}*\frac{1*10^{-9}C}{nC}=\frac{1.69N}{C}x\]
for q:
\[E_{q}=\frac{8.99*10^{9}Nm^{2}}{C^{2}}*\frac{-2nC}{29m^{2}}*\frac{1*10^{-9}C}{nC}=\frac{-0.62N}{C}\]
\[R=2mx+5my\]
\[r=\frac{R}{|R|}=\frac{2x}{\sqrt{29}}+\frac{5y}{\sqrt{29}}\]
\[\frac{-0.62N}{C}(\frac{2}{\sqrt{29}}x+\frac{5}{\sqrt{29}}y)\]

Answer 2

http://prntscr.com/9uf87w

Answer 3

the 2 and the 5?

Answer 4

ya or take the 0.62 as positive

Answer 5

Oh right let me fix that

Answer 6

\[E_{tot}=\frac{1.69N}{C}x+\frac{1.24N}{\sqrt{29}C}x+\frac{3.1N}{\sqrt{29}C}y\]
\[E_{tot}=\frac{1.69*\sqrt{29}N+1.24N}{\sqrt{29}C}x+\frac{3.1N}{\sqrt{29}C}y\]
\[E_{tot}=\frac{1.920N}{C}x+\frac{0.576N}{C}y\]

Answer 7

\[\sqrt{(1.920N/C)^{2}(0.576N/C)^{2}}=2N/C\]

Answer 8

k good (:

Answer 9

\[\theta=tan^{-1}(\frac{0.576}{1.920})=16.7\]

Answer 10

yep gj

Answer 11

thank you for the help =)

Answer 12

welcome!

Answer 13

b: Determine the magnitude and direction of the acceleration of a 2 µC charge of mass 10^−25kg which is released at the origin.

Answer 14

F=q*E

Answer 15

\[F=2\mu C*E=2\mu C*\frac{10^{-6}}{1\mu }*\frac{2N}C{}=4.0*10^{-6}N\]

Answer 16

\[F=m*a\]
\[\frac{F}{m}=a=\frac{4.0*10^{-6}N}{10^{-25}kg}=4.0*10^{19}\]

Answer 17

yep now apply it to the right direction

Answer 18

should be in the same direction as the field no?

Answer 19

yeah

Answer 20

sweet. thanks again

Answer 21

have u learnt about complex numbers yet

Answer 22

u dont need to do this, u can just find the x and y lengths for a 16.7 degree angle but

a shortcut since u know the angle already

\[Acceleration=exp(i*\frac{16.7}{360}*2\pi) ~*~4.0*10^{19}\]
the real part of that is in the x component of acc
and the imaginary part is the y component of acc

just ignore, if it it doesnt make sense. i just think its kind of a neat fast way to get vector representations of magnitudes in any direction quickly once u know the angle u want it to go in

Answer 23

or

Acc=4.0*10^19 * cos(16.7deg) x + 4.0*10^19 * sin(16.7deg) y