Ples help me?????????

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1 Answer
Feb 1, 2018

10 . Clearly the electron went opposite to the direction of the field vector,from negative to the positive plate due to an attractive force,so in this process it had undergone an acceleration due to the force generated due to the electric field.

So,if the electric field strength is #E# then force (#F#) generated due to this field is given as #F=Eq#(where, #q# is the charge of the electron.

So,acceleration caused is #F/m=(Eq)/m# where, #m # is the mass of electron.

As,the force acting is constant,so we can apply #v^2=u^2+2as#

Given, #v=5 *10^7 m/s,u=0,s=1.2/100,a=(Eq)/m#

So,solving the above,we get, #E=59.309 N/C#

9 . The charge lies at a distance (#r#) of #sqrt(0.2^2+0.3^2) m# or, #sqrt(0.13)#

So,force that will be acting on a point charge at origin,due to this #5 mu C# charge is #(9*10^9*5*10^-6*1/0.13) N# or #34.6*10^4 N#

So,electric field at origin will be #(F/q)# i.e #(34.6*10^4/1) N/C# (#q# is the point charge=#1 C#)

Similarly at point #(1,1)# , distance from the location of the #5muC# charge is #sqrt((1-0.2)^2+(1-0.3)^2)# i.e #sqrt(1.13)#

So,similarly electric field value will be #F/q# i.e #(9*10^9*5*10^-6*1/1.13)/1# i.e #39.82*10^3 N/C#