A charge of #-6 C# is at the origin. How much energy would be applied to or released from a # 9 C# charge if it is moved from # (5 ,-2 ) # to #(2 ,-1 ) #?

2 Answers
Jan 13, 2018

Answer:

(127.09)*(10^9) Joule amount of energy has to be taken out from the charge.

Explanation:

Initial potential energy of the charge is,
Ei = #(9*10^9)*{(-6)*9}/(√29) #[r=distance between them = #√{(5)^2+(-2)^2}#]
And final potential energy=
Ef = #(9*10^9)*{(-6)*9}/(√5) #[r1=distance between them =#√{(2)^2+(-1)^2}#
Hence change in energy =
Ef-Ei
= #(9*10^9){(-6)*9}{1/(√5)-1/(√29)}#

Or,-#(127.09)*(10^9)# Joule
Negative sign means energy has to be taken out from the charge

Jan 13, 2018

Answer:

The energy applied is #=127.1*10^9J#

Explanation:

The potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-6C#

The charge #q_2=9C#

The Coulomb's constant is #k=9*10^9Nm^2C^-2#

The distance

#r_1=sqrt((5)^2+(-2)^2)=sqrt(29)m#

The distance

#r_2=sqrt((2)^2+(1)^2)=sqrt(5)#

Therefore,

#U_1=k(q_1q_2)/r_1#

#U_2=k(q_1q_2)/r_2#

#DeltaU=U_2-U_1=k(q_1q_2)/r_2-k(q_1q_2)/r_1#

#=k(q_1q_2)(1/r_1-1/r_2)#

#=9*10^9*((-6)*(9))(1/sqrt29-1/sqrt(5))#

#=127.1*10^9J#