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Fundamentals Of Physics
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Short Answer

Figure 22-38a shows two charged particles fixed in place on an x-axis with separation L. The ratio q1/q2 of their charge magnitudes is . Figure 22-38b shows the x component Enet,Xof their net electric field along the x-axis just to the right of particle 2. The x-axis scale is set by xs=30.0 cm. (a) At what value of x>0 is Enet,x is maximum? (b) If particle 2 has charge -q2=-3e, what is the value of that maximum?

  1. The value of x at which Enet,Xis maximum is 34 cm
  2. The value of that maximum if particle 2 has a charge -3e is 2.2×10-8 N/C
See the step by step solution

Step by Step Solution

Step 1: The given data

  1. The two charged particles are on the x-axis with separation, L .
  2. The value of the ratio, q1q2=4
  3. The x-axis scale is set as: xs=30.0 cm
  4. Particle 2 has a charge of -q2=-3e

Step 2: Understanding the concept of electric field 

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again for the maximum value of the net field, we can differentiate the electric field equation for getting the value of x. Now, substituting the value of x, we can get the value of the required electric field.


The magnitude of the electric field, E=q4πε0R2R^ (1)

where R = The distance of field point from the charge, and q = charge of the particle

According to the superposition principle, the electric field at a point due to more than one charge,

E=i-1nEi =i-1nqi4πε0ri2r^i (2)

Step 3: a) Calculation of the value of Enetx for  being maximum

For it to be possible for the net field to vanish at some x > 0, the two individual fields (caused by q1and q2) must point in opposite directions for x > 0. They are therefore oppositely charged considering their positions. Further, since the net field points more strongly leftward for the small positive x (where it is very close to q2 ), then we conclude that localid="1657282744540" q2is the negative-valued charge. Thus, q1is a positive-valued charge.

From the given ratio, we can now considered for getting a maximum electric field that

q1=4e and q2 =e

Thus using equation (1), we can find the individual fields, and substituting this into equation (2), we can get the net electric as:

Enet=E1+E2 =44πε0(L+x)2-e4πε0(x)2...................(3)

Setting Enet=0 at (see graph) x = 20 cm , the graph immediately leads to To get the maximum value of the electric field, we can differentiate the above equation for x, and equating it to zero, we can get the value of x as:

ddx4e4πε0(L+x)2-e4πε0(x)2=0x=2323+1343+13Lx=1.70(20 cm)x=34 cm

Hence, the value of x is 34 cm

Step 4: b) Calculation of that maximum electric field

Substituting the given values in equation (3), we can the maximum electric field for the charge of particle 2, as follows:

E=4e4πε0(L+x)-3e4πε0(x)2 =9×109N.m2.C-2×1.6×10-19C4(0.34 m+0.20 m)2-3(0.20m)2 =2.2×10-8N/C

Hence, the value of the electric field is 2.2×10-8 N/C

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