25. Consider an infinitesimal section of the rod of length dx, a distance x from the left end, as shown in the

diagram below. It contains charge dq = λ dx and is a distance r from P . The magnitude of the field it
produces at P is given by

dE =

1

4πε

0

λ dx

r

2

.

The x component is

dE

x

=

−

1

4πε

0

λ dx

r

2

sin θ

and the y component is

dE

y

=

−

1

4πε

0

λ dx

r

2

cos θ .

x

y

dq

x

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R

r

θ

P

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.d 

E

We use θ as the variable of integration and substitute r = R/ cos θ, x = R tan θ and dx = (R/ cos

2

θ) dθ.

The limits of integration are 0 and π/2 rad. Thus,

E

x

=

−

λ

4πε

0

R

π/2

0

sin θ dθ =

λ

4πε

0

R

cos θ



π/2

0

=

−

λ

4πε

0

R

and

E

y

=

−

λ

4πε

0

R

π/2

0

cos θ dθ =

−

λ

4πε

0

R

sin θ



π/2

0

=

−

λ

4πε

0

R

.

We notice that E

x

= E

y

no matter what the value of R. Thus, 

E makes an angle of 45

◦

with the rod

for all values of R.