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# Charge and Field watch

1. Please could someone work out the answer to the electric potential in Q 4) c) - I got the rest of it right, but my answer was different to the book for that part
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2. (Original post by pippabethan)
Please could someone work out the answer to the electric potential in Q 4) c) - I got the rest of it right, but my answer was different to the book for that part
What did the book get, and what did you get?
3. (Original post by TheBride)
What did the book get, and what did you get?
I got 2500V, whereas the book got 5000V
4. (Original post by pippabethan)
Please could someone work out the answer to the electric potential in Q 4) c) - I got the rest of it right, but my answer was different to the book for that part
The Pd across the gap is 10KV. So halfway the PD will be half of that which is 5000V
5. (Original post by SirRaza97)
The Pd across the gap is 10KV. So halfway the PD will be half of that which is 5000V
Oh okay, thank you!
I used the electric field strength from V/d and multiplied by r to get Velec. Is this an incorrect method?
6. (Original post by pippabethan)
Oh okay, thank you!
I used the electric field strength from V/d and multiplied by r to get Velec. Is this an incorrect method?
Yes you can use that equation! Field Strength is the same but the distance r is half
7. (Original post by SirRaza97)
Yes you can use that equation! Field Strength is the same but the distance r is half
Aha maybe that's where I went wrong! Thank you

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