Two identical parallel plate capacitor are connected in series combination as shown in figure. Potential difference across $$capacitor1$$ is found to be $$4$$ volt. Now separation between the plates of capacitor $$2$$ is halved. Then new potential difference across $$1$$ will be

Question

Infinity Learn · Accepted Answer

$$c=Initial$$  capacitance of each capacitor. ∴$$Ce=c$$×$$cc+c=c/2$$ ∴ $$Q=$$ charge on each capacitor C.$$E=CE2$$∴$$QC=1C$$×$$CE2=4$$⇒$$E=8$$ voltFinally, $$c21=2c$$,   $$ce1=c$$×$$2cc+2c=2c3$$,   $$Q1=ce1E=2CE3$$∴ Final p.d. across $$1=Q1/C=2E3=2$$×$$83volt=5.33$$ volt

Two identical parallel plate capacitor are connected in series combination as shown in figure. Potential difference across capacitor1 is found to be 4 volt. Now separation between the plates of capacitor 2 is halved. Then new potential difference across 1 will be

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