IIT JEE , AIEEE Forum - Physics , Electricity

joyfrancis

Q1)Charge q is uniformly distributed over a thin half ring of radius R. The lectric field at the center of the ring is ?

Q2)Electric potential in an electric field is given by V=k/r.If position vector r=2i+3j+4k, then electric field will be ?

a5hw1n_5

the 1st answer should be

E= q/2

^2R^2

_o

the second i think it should be k/29.

plz correct me if wrong

neeraj_agarwal_1990

Use the formula.... Q                     sin(alpha/2)
                          -------------              ----------------
                       4pi(ep)R(sq)    alpha/2

alpha = pi

nick

lets consider an element whose charge is dq

dq= q/pi.r rd@ where @ is small angular displacement
r gets cancelled

now field due to small part is dE= dq/kr^2

now taking the components cos@ is cancelled by all the parts by symmetry
only sin@ will "add" up
hence
dEsin@=dq/kr^2 sin@=dEtotal

q/kpi.r^2 d@sin@
integrating from limits 0 to pi

we shall get
Etotal= 2q/pir^2.k here k=4pi.e' e'=epsilon

this is the ans..

elessar_iitkgp

Consider an element making an angle

with the diameter of the half ring and subtending d

at its center.
Charge on the element dq = (q/

R)Rd

= (q/

)d

Electric field at the center due to this element
dE = (1/4

)(dq/R²) = (1/4

)[(q/

)d

/ R²]
E =

dE sin

= (q/4

²

R²)₀

^pi sin

d

E = q/2

²

R²

kmr

electrostats hai naa...woh bahut mazedaar topic he hahaha

elessar_iitkgp

For the second part
E = - dV/dr = k/r²
For the position vector
r = 2i+3j+4k
r =

29
Hence at the point given
E = k/29

nnawandar

Q1} zero

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