Edit View History Window Help tritonphysicsndsunodakedu LONC

Edit View History Window Help triton,physics.ndsu.nodak.edu LON-CAPA Electric Flold of Two Particles Two Smal, Identical 01 72 91 In the figure, particle 1 of charge q1 3.00x10s C and particle 2 of charge 42 -1.50x1o C are fixed to an x axis, separated by a distance d - 0.100 m. Calculate their net electric field E(x) as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to the left. What is E(-O.100)? Submit Answer Tries o/s What is Ec-0.020)7 Submit Answor Tries o/s What is E(O.090)? Submit Anawer Tries 0/5 What is E(o.140)? Tries 0/5 Submit Answor Export NEW CLARIFICATION John Buncher Reply (Mon Jan 18 10:26:21 pm 2016 (CST) ether restate email or phone -your preferenice

Solution

For points lying to the left of origin, the electric fields due to the individual charges will add up,

Hence for a negative x, we have Enet = -K(q1/r1^2 + q2/r2^2)= -8.99 x 10^4 [3 / x*2 + 15/ (x+0.1) ^2]

For x to the right of x =d, we have Enet = K(q1/r1^2 + q2/r2^2)= 8.99 x 10^4 [3/ x*2 + 15/ (x-0.1) ^2]

For x between x= 0 and x = d, Enet = K(q1/r1^2 - q2/r2^2)= 8.99 x 10^4 [3/ x*2 - 15/ (0.1 - x) ^2]

Now we put in the values appropriately as:

E(-0.1) = -8.99 x 10^4 [300 + 375] = -6.068 x 10^7 N/C

E(-0.02) = - 8.99 x 10^4 [7500 + 1041.67] = -7.6789 x 10^8 N/C

E(0.09) = 8.99 x 10^4 [3 / 0.09*0.09 - 15 / 0.01*0.01 ] = -1.34517 x 10^10 N/C

E(0.140) = 8.99 x 10^4 [3 / 0.14*0.14 + 15 / 0.04*0.04] = 8.5657 x 10^8 N/C