Two solid spheres carrying equal charges have radii 5cm and 7cm respectively .the ratio of there respective electric fields
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As we know ,
Electric field (electric field strength) of solid sphere, E= (1/(4π€))*q/r^2
where q= charge carring by solid sphere,
r=radius of the sphere,
And (1/(4π€))= Coulomb's constant= 9*10^9 Nm^2/C^2.
Given ,
Radius, r1= 5 cm,
Radius, r2= 7cm,
As the both sphere carry equal charge i.e., q , it is considered as constant here.
Let electric field of two spheres be E1 and E2 respectively
Then the ratio of electric field, E1/E2= [(1/(4π€))*q/r1^2]/[(1/(4π€))*q/r2^2]
=[1/r1^2]/[1/r2^2] ;[as constant terms are cancelled out]
=r2^2/r1^2
=(7cm)^2/(5cm)^2
=49/25
Therefore the ratio of electric field E1/E2= 49/25. (Ans.)
Electric field (electric field strength) of solid sphere, E= (1/(4π€))*q/r^2
where q= charge carring by solid sphere,
r=radius of the sphere,
And (1/(4π€))= Coulomb's constant= 9*10^9 Nm^2/C^2.
Given ,
Radius, r1= 5 cm,
Radius, r2= 7cm,
As the both sphere carry equal charge i.e., q , it is considered as constant here.
Let electric field of two spheres be E1 and E2 respectively
Then the ratio of electric field, E1/E2= [(1/(4π€))*q/r1^2]/[(1/(4π€))*q/r2^2]
=[1/r1^2]/[1/r2^2] ;[as constant terms are cancelled out]
=r2^2/r1^2
=(7cm)^2/(5cm)^2
=49/25
Therefore the ratio of electric field E1/E2= 49/25. (Ans.)
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