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A charge of #7 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, 1 ) # to #(2 ,-1 ) #?

Answer 1

Sebastian Acosta

The energy applied is #=142.4*10^9J#

Potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=7C#

The charge #q_2=9C#

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

The separation

#r_1=sqrt((-5)^2+(1)^2)=sqrt26m#

The separation

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

Consequently,

#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*((7)*(9))(1/sqrt5-1/(sqrt26))#

#=142.4*10^9J#

The energy released is #=142.4*10^9J#

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Answer 2

Layla Ahmed

The energy applied to the 9 C charge would be 162 joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.