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Effect of radius changes on electric fields and potential difference? 
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#1
Jan3112, 01:21 AM

P: 40

Hello! Okay so I understand that electric potential: V = kQ/r ...must be influenced by the radius doubling because it would make the potential energy half of what it originally was because of the proportionality law, v is proportional to 1/r. With electric fields though, how can there possibly be no change? The formula is E = kQ/r^2 ...meaning if should be influenced as well? The answer is c, btw. 


#2
Jan3112, 05:39 AM

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P: 12,974

So do the math  calculate the field before and after the change in position.
Note: E is a vector. 


#3
Jan3112, 08:46 AM

P: 40

Oh! Right!
Since E is a vector, and the distance and charges on both sides are equal, then they always simply cancel out to zero. Whereas with electric potential, a scalar quantity, it is not affected by direction, merely magnitude, in which case both are positive, so the radius change will definitely affect it. Thanks bud! 


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