Dielectric Breakdown

[sshah92]

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From the TPR book:

"The presence of a dielectric increases the maximum electric field strength. As a result, capacitors with dielectrics can hold more charge (and thus store more potential energy) without the threat of dielectric breakdown."

I'm a little confused because as I understood it, there are two ways of inserting a dielectric. One is by charging the capacitor with a battery and inserting the dielectric before removing the battery from the circuit. Q = C*V so the capacitance increases by a factor of K, but the voltage stays the same (same electric potential difference between terminals), so Q increases. But if V stays the same, and because V = E*d, the electric field stays the same.

So why would the electric field strength increase? Is there a difference between electric field and electric field strength?

(The other way is by charging the capacitor with the battery, removing the battery, and then inserting the dielectric. This takes work that is being done against the electric field. And without a long-winded explanation, E ultimately decreases in this case by a factor of K. So I doubt this case is the one being referred to in this passage.)

Please help! Sorry for asking all this questions today haha, electrostatics is one of my weaker areas in PS 🙁

 

[pm1]

From the TPR book:

"The presence of a dielectric increases the maximum electric field strength. As a result, capacitors with dielectrics can hold more charge (and thus store more potential energy) without the threat of dielectric breakdown."

I'm a little confused because as I understood it, there are two ways of inserting a dielectric. One is by charging the capacitor with a battery and inserting the dielectric before removing the battery from the circuit. Q = C*V so the capacitance increases by a factor of K, but the voltage stays the same (same electric potential difference between terminals), so Q increases. But if V stays the same, and because V = E*d, the electric field stays the same.

So why would the electric field strength increase? Is there a difference between electric field and electric field strength?

(The other way is by charging the capacitor with the battery, removing the battery, and then inserting the dielectric. This takes work that is being done against the electric field. And without a long-winded explanation, E ultimately decreases in this case by a factor of K. So I doubt this case is the one being referred to in this passage.)

Please help! Sorry for asking all this questions today haha, electrostatics is one of my weaker areas in PS 🙁

Yes, I agree with you that they are probably talking about the case where the battery is still attached. This way, voltage remains the same. C=Q/V. C always increases when added a dielectric. So with V remaining the same Q must increase.

As you mentioned, because V stays the same E must also stay the same (according to V = Ed).

However, F = Eq. Since q increased F will also increase even though E remained the same.

(Please correct if I am wrong. Electrostatics are not my cup of tea either.. hehe I will gladly talk about it to get more practice! 🙂

 

[sshah92]

Yes, I agree with you that they are probably talking about the case where the battery is still attached. This way, voltage remains the same. C=Q/V. C always increases when added a dielectric. So with V remaining the same Q must increase.

As you mentioned, because V stays the same E must also stay the same (according to V = Ed).

However, F = Eq. Since q increased F will also increase even though E remained the same.

(Please correct if I am wrong. Electrostatics are not my cup of tea either.. hehe I will gladly talk about it to get more practice! 🙂

Ohhh. So the quote referred to force instead of field. That makes so much more sense! Thanks haha