A 1.88 10-9 C charge has coordinates x = 0, y = ?2.00; a 2.76 10-9 C charge has
ID: 2162604 • Letter: A
Question
A 1.88 10-9 C charge has coordinates x = 0, y = ?2.00; a 2.76 10-9 C charge has coordinates x = 3.00, y = 0; and a -5.15 10-9 C charge has coordinates x = 3.00, y = 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin.(a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis).
magnitude
direction
(b) Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis).
magnitude
direction
Explanation / Answer
Just calculate the e-field vector caused by each of the charges by itself (as if none of the other charges were present), and then add them all up. It really is that simple plug your take help Distance of q1 from origin=r1= sq rt(0.6)^2 + (0.8)^2=1 m Distance of q2 from origin=r2=0.6 m Electric field due to q1 =E1=kq1/r1^2 Electric field due to q1 =E1=9*10^9*4*10^-9/1^2 Electric field due to q1 =E1=36 N/C at angleO with x axis,where tanO=53 degree Component of E1 along x axis =36cos53=36*0.6=21.6 N/C (i) Component of E1 along y axis =36sin53=36*0.8=28.8 N/C (j)_______________________________ Electric field due to q2 =E2=kq2/r2^2 Electric field due to q2 =E2=9*10^9*6*10^-9/0.6^2 Electric field due to q2 =E2=150 N/C along (-) x axis In vector notation,E2 =150N/C(-i) _____________________________ Net field = E = E1+ E2 E = 21.6 N/C (i) + 28.8 N/C (j) + 150N/C(-i) E = 128.4 N/C (-i) + 28.8 N/C(j) Magnitude of resultant = E =sq rt 17316 =131.59 N/C Angle O which Emakes with negative x axis is given by tanO= 28.8/128.4=0.22429 Angle O =12.64 degree clockwise with negative x axis or 192.64 degree clockwise with positive x axis
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