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Two particles, each of positive charge q, are fixed in place on a yaxis, one aty=dand the other at.y=-d (a) Write an expression that gives the magnitude Eof the net electric field at points on the xaxis given by.x=ad (b) Graph Eversusfor the range0<<4. From the graph, determine the values ofthat give (c) the maximum value of Eand (d) half the maximum value of E.

Short Answer

Expert verified

a) The expression for the magnitude of the electric field on the x-axis is q4od2((2+1)3/2).

b) The graph of the electric field versus is plotted.

c) The value of for the maximum value of the electric field, E is1/鈭 2.

d) The values of for half the maximum value of E are 0.2047 and 1.9864.

Step by step solution

01

The given data

Two particles, each of positive charge, are fixed in place on a y axis, one at y=dand the other aty=-d.

02

Understanding the concept of electrostatic force 

Using the concept of the electric field, we can get the required expression of the field. The value of the distance t which we get the maximum field is calculated by equating the differentiation of the electric field expression to zero. Similarly, the value of the distance is calculated for half the maximum electric field.

Formula:

The magnitude of the electric field, E=q4or2r^where,r^=cosi^+sinj^ (i)

where, r = The distance of field point from the charge

q = charge of the particle

03

a) Calculation of the expression of the electric field

Letq1denote the charge aty=dandq2denote the charge at.y=d The individual magnitudesE1andE2are figured from equation (i), where the absolute value signs forqare unnecessary since these charges are both positive. The distance fromq1to a point on the x-axis is the same as the distance fromto a point on the x-axis, which is given as:

r=(x2+d2)

By symmetry, the y-component of the net field along the x axis is zero.

Thus, the x component of the net field, evaluated at points on the positive x-axis, is given as:

Ex=2(14o)(qx2+d2)(xx2+d2)

Where the last factor iscos=x/rwithbeing the angle for each individual field as measured from the x axis.

Solving the above expression, by substituting the value,x=d we obtain the electric field as:

Ex=q4od2((2+1)3/2)

Hence, the value of the expression of the electric field is.q4od2((2+1)3/2)

04

b) Calculation for plotting the graph of electric field

The graph ofE=Exversusis shown below. For the purposes of graphing, we set d=1mand.q=5.56x1011C

Hence, the required graph is plotted.

05

c) Calculation of the value of for maximum electric field

From the graph, we estimate the electric fieldEmaxoccurs at=0.71about.

Thus, the value at which we get the maximum field occur is1/鈭 2.

06

d) Calculation of the value of for half of the maximum electric field

The graph suggests that 鈥渉alf-height鈥 points occur at0.2and2.0 .

Further numerical exploration the desired value of the electric field leads to the values of as 0.2047 and 1.9864.

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